OP Malhotra Class 12 Maths Solutions Chapter 17 Differential Equations Exercise 17 (D)

S Chand Class 12 ICSE Maths Solutions Chapter 17 Differential Equations Ex 17(D)

Solve the Following Differential Equations:

Question 1.
(i) \( \frac { dy }{ dx } = e^{x+y} \)
(ii) \( \frac { dy }{ dx } = x³ e^{-2y} \)
Answer:
(i) The given equation is \( \frac { dy }{ dx } = e^{x+y} \). This can be written as \( \frac { dy }{ dx } = e^x e^y \). We separate the variables, putting terms with \( y \) on one side and terms with \( x \) on the other. This gives us \( e^{-y} dy = e^x dx \). Next, we integrate both sides. The integral of \( e^{-y} \) is \( -e^{-y} \), and the integral of \( e^x \) is \( e^x \). So, we get \( -e^{-y} = e^x + c \), where \( c \) is the integration constant. Rearranging the terms, we find the required solution: \( e^x + e^{-y} = C \). This type of equation shows how two changing quantities are related through their rates of change.
(ii) The given differential equation is \( \frac { dy }{ dx } = x³ e^{-2y} \). First, we separate the variables so all \( y \) terms are with \( dy \) and all \( x \) terms are with \( dx \). This gives us \( e^{2y} dy = x³ dx \). Then, we integrate both sides. The integral of \( e^{2y} \) is \( \frac{e^{2y}}{2} \) and the integral of \( x³ \) is \( \frac{x^4}{4} \). After integrating and adding the constant \( C \), we get \( \frac{e^{2y}}{2} = \frac{x^4}{4} + \frac{C}{4} \). Multiplying by 4 to clear fractions, we arrive at the final solution: \( 2e^{2y} = x^4 + C \). This process helps solve problems where the rate of change depends on both variables.
In simple words: For (i), separate 'y' and 'x' terms, then integrate each side. Rearrange for the final form. For (ii), separate 'y' and 'x' terms, then integrate each side. Multiply by 4 to remove fractions and find the final answer.

🎯 Exam Tip: Always remember to add the constant of integration (C) after integrating and simplify the expression for a clean final solution.

Question 2.
\( \frac{d y}{d x}=\frac{x y+y}{xy+x} \)
Answer: The given differential equation is \( \frac{d y}{d x}=\frac{x y+y}{xy+x} \). First, we factor out common terms from the numerator and denominator to get \( \frac{d y}{d x}=\frac{y(x+1)}{x(y+1)} \). Next, we separate the variables, grouping all \( y \) terms with \( dy \) and all \( x \) terms with \( dx \), which results in \( \frac{y+1}{y} dy = \frac{x+1}{x} dx \). We can rewrite this as \( \left(1+\frac{1}{y}\right) dy = \left(1+\frac{1}{x}\right) dx \). Now, we integrate both sides. The integral of \( 1 \) is \( y \) or \( x \), and the integral of \( \frac{1}{y} \) or \( \frac{1}{x} \) is \( \log|y| \) or \( \log|x| \). So, the solution becomes \( y + \log|y| = x + \log|x| + C \). This technique is very useful for solving differential equations that can be broken into parts.
In simple words: Factor out common parts. Put 'y' terms with 'dy' and 'x' terms with 'dx'. Integrate both sides. This gives the answer with 'y', 'x', and 'log' terms.

🎯 Exam Tip: Factoring the numerator and denominator correctly is the crucial first step to identify if the differential equation is separable.

Question 3.
\( y (1 - x²) dy = x(1 + y²) dx \)
Answer: We start with the equation \( y (1 - x²) dy = x(1 + y²) dx \). To solve this, we first separate the variables. We move all \( y \) terms to the left side with \( dy \) and all \( x \) terms to the right side with \( dx \). This gives us \( \frac{y dy}{1+y^2} = \frac{x dx}{1-x^2} \). Next, we integrate both sides. We use the formula that the integral of \( \frac{f'(x)}{f(x)} \) is \( \log|f(x)| \). After adjusting the numerators by multiplying by 2 and -2 respectively (and compensating with \( \frac{1}{2} \) and \( -\frac{1}{2} \) outside the integral), we get \( \frac{1}{2} \log(1+y^2) = -\frac{1}{2} \log(1-x^2) + \frac{1}{2} \log C \). We then multiply by 2 and combine the log terms using log properties (\( \log A + \log B = \log(AB) \)). This simplifies to \( \log((1+y^2)(1-x^2)) = \log C \). Finally, we remove the log to get the solution: \( (1+y^2)(1-x^2) = C \). This method is good for equations where one side is a derivative of the other.
In simple words: Separate the 'y' and 'x' parts. Make the top part of each fraction the derivative of its bottom part. Integrate to get 'log' terms. Combine the 'log' terms and remove 'log' to find the answer.

🎯 Exam Tip: When integrating functions like \( \frac{x}{1-x^2} \), remember to adjust the numerator to be the exact derivative of the denominator (e.g., \( -2x \)) and compensate with a constant multiplier (e.g., \( -\frac{1}{2} \)).

Question 4.
\( \frac{d y}{d x}+\frac{\cos x \sin y}{\cos y} = 0 \)
Answer: The given equation is \( \frac{d y}{d x}+\frac{\cos x \sin y}{\cos y} = 0 \). First, we move the second term to the right side: \( \frac{d y}{d x} = -\frac{\cos x \sin y}{\cos y} \). Then, we separate variables, getting all \( y \) terms on the left and all \( x \) terms on the right. This gives us \( \frac{\cos y}{\sin y} dy = -\cos x dx \). Notice that \( \frac{\cos y}{\sin y} \) is \( \cot y \). Now, we integrate both sides. The integral of \( \cot y \) is \( \log|\sin y| \), and the integral of \( -\cos x \) is \( -\sin x \). So, we have \( \log|\sin y| = -\sin x + c \). We can rearrange this to \( \log|\sin y| + \sin x = c \). To remove the logarithm, we take \( e \) to the power of both sides, giving \( e^{\log|\sin y| + \sin x} = e^c \). This simplifies to \( \sin y e^{\sin x} = A \), where \( A \) is a new constant \( e^c \). Finally, we can write it as \( \sin y = Ae^{-\sin x} \). This is a common way to solve equations where variables can be grouped.
In simple words: Move terms to separate 'y' and 'x'. Integrate the 'cot y' part and the 'cos x' part. Use logarithms to simplify. The final answer will show 'sin y' and 'sin x' linked by a constant.

🎯 Exam Tip: When dealing with logarithmic constants, it's often helpful to express the final constant as \( e^c \) (or A) for a cleaner solution, especially after taking exponentials of both sides.

Question 5.
(i) \( (y + xy) dx + (x - xy²) dy = 0 \)
(ii) \( (x² - yx²) dy + (y² + xy²) dx = 0 \)
(iii) \( x² (y + 1) dx + y² (x- 1) dy = 0 \)
Answer:
(i) We begin with the differential equation \( (y + xy) dx + (x - xy²) dy = 0 \). First, we factor out common terms, which gives us \( y(1 + x) dx + x(1 - y²) dy = 0 \). To separate the variables, we divide the entire equation by \( xy \). This makes the equation \( \frac{1+x}{x} dx + \frac{1-y^2}{y} dy = 0 \). We can simplify each fraction: \( \left(\frac{1}{x}+1\right) dx + \left(\frac{1}{y}-y\right) dy = 0 \). Now, we integrate each part separately. The integral of \( \frac{1}{x} \) is \( \log|x| \), the integral of \( 1 \) is \( x \), the integral of \( \frac{1}{y} \) is \( \log|y| \), and the integral of \( -y \) is \( -\frac{y²}{2} \). Adding an integration constant \( c \), the solution is \( \log|x| + x + \log|y| - \frac{y²}{2} = c \). This method is effective for equations where terms can be separated and integrated independently.
(ii) We are given the equation \( (x² - yx²) dy + (y² + xy²) dx = 0 \). First, we factor out common terms: \( x²(1 - y) dy + y²(1 + x) dx = 0 \). To separate variables, we divide the entire equation by \( x²y² \), yielding \( \frac{1-y}{y^2} dy + \frac{1+x}{x^2} dx = 0 \). We can split these fractions into simpler terms: \( \left(\frac{1}{y^2}-\frac{1}{y}\right) dy + \left(\frac{1}{x^2}+\frac{1}{x}\right) dx = 0 \). Now, we integrate each term. The integral of \( \frac{1}{y^2} \) is \( -\frac{1}{y} \), \( -\frac{1}{y} \) is \( -\log|y| \), \( \frac{1}{x^2} \) is \( -\frac{1}{x} \), and \( \frac{1}{x} \) is \( \log|x| \). Combining these with a constant \( c \), we get \( -\frac{1}{y} - \log|y| - \frac{1}{x} + \log|x| = c \). Rearranging the terms, the solution can be written as \( \log\left|\frac{x}{y}\right| = \frac{x+y}{xy} + c \). Separating variables is a fundamental method in solving these equations.
(iii) The equation is \( x² (y + 1) dx + y² (x - 1) dy = 0 \). We first separate the variables by dividing by \( (x-1)(y+1) \), which gives \( \frac{x²}{x-1} dx + \frac{y²}{y+1} dy = 0 \). To integrate these, we perform polynomial division on each fraction. \( \frac{x²}{x-1} \) becomes \( x+1+\frac{1}{x-1} \), and \( \frac{y²}{y+1} \) becomes \( y-1+\frac{1}{y+1} \). Now, we integrate each term. The integral of \( x \) is \( \frac{x²}{2} \), \( 1 \) is \( x \), \( \frac{1}{x-1} \) is \( \log|x-1| \). Similarly for \( y \). Combining these, we get \( \frac{x²}{2} + x + \log|x-1| + \frac{y²}{2} - y + \log|y+1| = c \). This solution is obtained by carefully separating terms and integrating them.
In simple words: For each part, first factor out common terms, then separate 'x' and 'y' parts. For (i) and (ii), divide by 'xy' or 'x²y²' and split fractions. For (iii), divide fractions using polynomial division. Integrate each simple part to get the final answer with 'log' and power terms.

🎯 Exam Tip: Always look for common factors to simplify the equation before attempting to separate variables. Polynomial division is useful for integrating rational functions where the numerator's degree is greater than or equal to the denominator's degree.

Question 6.
\( \sqrt{a+x} \frac{d y}{d x} = - xy \)
Answer: We start with the differential equation \( \sqrt{a+x} \frac{d y}{d x} = - xy \). First, we separate the variables, putting \( y \) terms on one side and \( x \) terms on the other: \( \frac{dy}{y} = \frac{-x}{\sqrt{a+x}} dx \). Now, we integrate both sides. The left side is a simple integral: \( \int \frac{dy}{y} = \log|y| \). For the right side, we use a substitution: let \( u = a+x \). This changes the integral into \( \int \frac{-(u-a)}{\sqrt{u}} du \), which can be split into \( \int (-u^{1/2} + au^{-1/2}) du \). Integrating this gives \( -\frac{2}{3}u^{3/2} + 2a u^{1/2} \). Substituting back \( u = a+x \), we get \( -\frac{2}{3}(a+x)^{3/2} + 2a(a+x)^{1/2} \). So, the final solution is \( \log|y| = -\frac{2}{3}(a+x)^{3/2} + 2a\sqrt{a+x} + c \). This method uses substitution to simplify complex integrals.
In simple words: Separate 'y' terms and 'x' terms. Integrate the 'dy/y' part to get 'log y'. For the 'x' part, use a substitution (like \( u = a+x \)) to make it simpler to integrate. After integrating, change 'u' back to 'x' terms to find the full answer.

🎯 Exam Tip: For integrals involving square roots of linear expressions like \( \sqrt{ax+b} \), substitution of \( u = ax+b \) is usually the most effective approach.

Question 7.
\( \frac { dy }{ dx } = 1 - x + y - xy \)
Answer: We are given the differential equation \( \frac { dy }{ dx } = 1 - x + y - xy \). First, we factor the right-hand side by grouping terms. This results in \( \frac { dy }{ dx } = (1-x)(1+y) \). Then, we separate the variables, placing all \( y \) terms with \( dy \) and all \( x \) terms with \( dx \). This gives us \( \frac{dy}{1+y} = (1-x) dx \). Next, we integrate both sides. The integral of \( \frac{1}{1+y} \) is \( \log|1+y| \), and the integral of \( (1-x) \) is \( x - \frac{x²}{2} \). Adding an integration constant \( c \), the solution is \( \log|1+y| = x - \frac{x²}{2} + c \). Factoring and separating variables are key steps to solve such equations.
In simple words: First, simplify the right side by factoring it. Then, separate the 'y' and 'x' terms to different sides. Integrate each side. You will get 'log' on one side and 'x' terms on the other, plus a constant.

🎯 Exam Tip: Always try to factor expressions on the right-hand side to check if the differential equation can be separated into functions of x and y.

Question 8.
\( (1 + x) (1 + y²) dx + (1 + y) (1 + x²) dy = 0 \)
Answer: We are given the equation \( (1 + x) (1 + y²) dx + (1 + y) (1 + x²) dy = 0 \). To solve it, we separate the variables by dividing the entire equation by \( (1+x²)(1+y²) \). This gives us \( \frac{1+x}{1+x²} dx + \frac{1+y}{1+y²} dy = 0 \). We split each fraction into two parts. The integrals for terms like \( \frac{1}{1+x²} \) are \( \tan^{-1}x \), and for terms like \( \frac{x}{1+x²} \) are \( \frac{1}{2}\log(1+x²) \). After integrating both sides and combining the terms, we get \( \tan^{-1}x + \frac{1}{2}\log(1+x²) + \tan^{-1}y + \frac{1}{2}\log(1+y²) = c \). This can be further simplified using the sum of inverse tangent formula and logarithm properties to \( \tan^{-1}\left(\frac{x+y}{1-xy}\right) + \frac{1}{2}\log((1+x²)(1+y²)) = c \). This solution involves using standard integral formulas for inverse tangents and logarithms.
In simple words: Separate 'x' and 'y' terms. Break down each fraction into two parts. Integrate each part separately. This will give 'inverse tan' and 'log' terms. Combine these using math rules to get the final answer.

🎯 Exam Tip: Remember standard integral forms like \( \int \frac{1}{1+x^2} dx = \tan^{-1}x \) and \( \int \frac{x}{1+x^2} dx = \frac{1}{2}\log(1+x^2) \) as they frequently appear in such problems.

Question 9.
\( \sec² x \tan y dx - \sec²y \tan x dy = 0 \)
Answer: We are given the differential equation \( \sec² x \tan y dx - \sec²y \tan x dy = 0 \). To solve this, we first separate the variables by dividing the entire equation by \( \tan x \tan y \). This yields \( \frac{\sec² x}{\tan x} dx - \frac{\sec² y}{\tan y} dy = 0 \). Next, we integrate both sides. We know that \( \sec² x \) is the derivative of \( \tan x \), so the integral of \( \frac{\sec² x}{\tan x} \) is \( \log|\tan x| \). Similarly, the integral for the \( y \) term is \( \log|\tan y| \). We add an integration constant, represented as \( \log C \). So, we have \( \log|\tan x| - \log|\tan y| = \log C \). Using logarithm properties, we combine the terms: \( \log\left|\frac{\tan x}{\tan y}\right| = \log C \). Removing the logarithms from both sides gives us \( \frac{\tan x}{\tan y} = C \). We can write this as \( \tan x = A \tan y \), where \( A \) is a constant. This problem shows how to integrate functions where the numerator is the derivative of the denominator.
In simple words: Separate the 'x' terms with 'dx' and 'y' terms with 'dy'. Notice that the top part of each fraction is the derivative of the bottom part. Integrate to get 'log' terms. Combine the 'log' terms and remove 'log' to find the final answer.

🎯 Exam Tip: Recognizing the \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| + C \) pattern is a shortcut that saves time when integrating. Make sure to choose the constant of integration in a form that simplifies the final expression.

Question 10.
\( \frac{d y}{d x}=e^{x-y}+e^{2 \log x-y} \)
Answer: The given differential equation is \( \frac{d y}{d x}=e^{x-y}+e^{2 \log x-y} \). First, we simplify the terms using exponent rules: \( e^{x-y} \) becomes \( e^x e^{-y} \), and \( e^{2 \log x-y} \) becomes \( e^{\log x²} e^{-y} \), which simplifies to \( x² e^{-y} \). So, the equation is \( \frac{d y}{d x}=e^x e^{-y} + x² e^{-y} \). We then factor out \( e^{-y} \) from the right side, giving \( \frac{d y}{d x}=e^{-y}(e^x + x²) \). Now, we separate the variables, moving \( e^{-y} \) to the left side: \( e^y dy = (e^x + x²) dx \). Finally, we integrate both sides. The integral of \( e^y \) is \( e^y \), and the integral of \( (e^x + x²) \) is \( e^x + \frac{x³}{3} \). Adding an integration constant \( c \), the solution is \( e^y = e^x + \frac{x³}{3} + c \). This problem shows how to simplify exponential and logarithmic terms before solving.
In simple words: Simplify the exponential terms using rules. Factor out the common part. Separate 'y' terms and 'x' terms. Integrate each side to get the answer with 'e' and 'x³' terms.

🎯 Exam Tip: Always simplify exponential and logarithmic expressions using their properties (e.g., \( e^{A+B} = e^A e^B \) and \( e^{k \log x} = x^k \)) before attempting variable separation.

Question 11.
(i) \( \cos x \cos y dy + \sin x \sin y dx = 0 \)
(ii) \( (1 + \cos x) dy = (1 - \cos y) dx \)
Answer:
(i) The given equation is \( \cos x \cos y dy + \sin x \sin y dx = 0 \). First, we separate variables by dividing by \( \cos x \sin y \). This yields \( \frac{\cos y}{\sin y} dy + \frac{\sin x}{\cos x} dx = 0 \). We know that \( \frac{\cos y}{\sin y} \) is \( \cot y \) and \( \frac{\sin x}{\cos x} \) is \( \tan x \). Now, we integrate both sides. The integral of \( \cot y \) is \( \log|\sin y| \), and the integral of \( \tan x \) is \( -\log|\cos x| \). We use \( \log C \) as the integration constant. So, we have \( \log|\sin y| - \log|\cos x| = \log C \). Using logarithm properties, we combine the terms: \( \log\left|\frac{\sin y}{\cos x}\right| = \log C \). Removing the logarithms from both sides gives us \( \frac{\sin y}{\cos x} = C \). Finally, we can write this as \( \sin y = A \cos x \), where \( A \) is a new constant. Separating variables and using known integral forms is key here.
(ii) We begin with \( (1 + \cos x) dy = (1 - \cos y) dx \). First, we separate the variables, putting \( y \) terms on the left and \( x \) terms on the right: \( \frac{dy}{1-\cos y} = \frac{dx}{1+\cos x} \). Next, we use the half-angle trigonometric identities: \( 1-\cos y = 2\sin²\left(\frac{y}{2}\right) \) and \( 1+\cos x = 2\cos²\left(\frac{x}{2}\right) \). Substituting these into the equation, we get \( \frac{dy}{2\sin²\left(\frac{y}{2}\right)} = \frac{dx}{2\cos²\left(\frac{x}{2}\right)} \). This can be rewritten using cosecant and secant functions: \( \frac{1}{2} \csc²\left(\frac{y}{2}\right) dy = \frac{1}{2} \sec²\left(\frac{x}{2}\right) dx \). Now, we integrate both sides. The integral of \( \csc²(Ay) \) is \( -\frac{1}{A} \cot(Ay) \), and the integral of \( \sec²(Ax) \) is \( \frac{1}{A} \tan(Ax) \). For our case, \( A = \frac{1}{2} \). This leads to \( -\cot\left(\frac{y}{2}\right) = \tan\left(\frac{x}{2}\right) + C \). Trigonometric identities are often essential for simplifying these differential equations.
In simple words: For (i), separate 'x' and 'y' terms. Integrate 'cot y' and 'tan x'. Use 'log' terms for the result. Combine 'log' terms and remove 'log' to get 'sin y' and 'cos x' linked by a constant. For (ii), separate 'y' terms and 'x' terms. Use special trigonometry rules (half-angle formulas) to change the 'cos' terms. Rewrite using 'cosec²' and 'sec²'. Integrate both sides to get the 'cot' and 'tan' terms with a constant.

🎯 Exam Tip: Keep a list of common trigonometric identities handy, especially half-angle formulas for \( 1 \pm \cos \theta \), as they are very useful in simplifying separable differential equations.

Question 12.
\( (1 + x²)dy + x\sqrt{1-y^2} dx = 0 \)
Answer: We are given the differential equation \( (1 + x²)dy + x\sqrt{1-y^2} dx = 0 \). First, we separate the variables by dividing the entire equation by \( (1+x²)\sqrt{1-y^2} \). This gives us \( \frac{dy}{\sqrt{1-y^2}} + \frac{x}{1+x²} dx = 0 \). Now, we integrate both sides. The integral of \( \frac{dy}{\sqrt{1-y^2}} \) is \( \sin^{-1}y \). For the second term, \( \int \frac{x}{1+x²} dx \), we can make the numerator the derivative of the denominator by multiplying and dividing by 2: \( \frac{1}{2}\int \frac{2x}{1+x²} dx \). The integral of this form is \( \frac{1}{2}\log(1+x²) \). Adding an integration constant \( c \), the solution is \( \sin^{-1}y + \frac{1}{2}\log(1+x²) = c \). Recognizing standard integral forms is crucial in solving such problems.
In simple words: Separate 'y' terms and 'x' terms. Integrate the 'y' part to get 'inverse sin'. For the 'x' part, make the top a derivative of the bottom (by adding 2 and 1/2). Integrate to get 'log' terms. Add a constant to find the answer.

🎯 Exam Tip: Memorize standard inverse trigonometric integral forms, such as \( \int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1}x \), as they frequently appear in variable separable equations.

Question 13.
\( (1 - x²) dy + xy dx = xy² dx \)
Answer: We start with the equation \( (1 - x²) dy + xy dx = xy² dx \). First, we rearrange the terms to group \( dx \) terms: \( (1 - x²) dy = xy² dx - xy dx \). This simplifies to \( (1 - x²) dy = xy(y - 1) dx \). Now, we separate the variables by dividing by \( y(y-1) \) and \( (1-x²) \), giving \( \frac{dy}{y(y-1)} = \frac{x}{1-x²} dx \). For the left side, we use partial fraction decomposition, rewriting \( \frac{1}{y(y-1)} \) as \( \frac{1}{y-1} - \frac{1}{y} \). For the right side, we adjust the numerator to be the derivative of the denominator (by multiplying by -2 and \( -\frac{1}{2} \)). So, we integrate \( \int \left(\frac{1}{y-1} - \frac{1}{y}\right) dy = -\frac{1}{2} \int \frac{-2x}{1-x²} dx \). This results in \( \log|y-1| - \log|y| = -\frac{1}{2} \log|1-x²| + c \). Combining the logarithm terms gives the final solution: \( \log\left|\frac{y-1}{y}\right| + \frac{1}{2} \log|1-x²| = c \). Partial fractions are a powerful tool for integrating rational functions.
In simple words: Move terms around to separate 'y' with 'dy' and 'x' with 'dx'. Use partial fractions to simplify the 'y' side. For the 'x' side, make the top the derivative of the bottom. Integrate both sides to get 'log' terms. Combine the 'log' terms for the final answer.

🎯 Exam Tip: Partial fraction decomposition is often needed for integrating rational functions after separating variables. Remember that \( \frac{1}{y(y-1)} \) can be decomposed into \( \frac{1}{y-1} - \frac{1}{y} \).

Question 14.
\( x\sqrt{1+y^2} d x+y \sqrt{1+x^2}dy = 0 \)
Answer: We are given the differential equation \( x\sqrt{1+y^2} d x+y \sqrt{1+x^2}dy = 0 \). The first step is to separate the variables. We divide the entire equation by \( \sqrt{1+x^2}\sqrt{1+y^2} \). This gives us \( \frac{x}{\sqrt{1+x^2}} dx + \frac{y}{\sqrt{1+y^2}} dy = 0 \). Now, we integrate both sides. To integrate terms like \( \int \frac{x}{\sqrt{1+x^2}} dx \), we use a substitution method. Let \( u = 1+x² \), then \( du = 2x dx \). This changes the integral to \( \int \frac{1}{2} \frac{du}{\sqrt{u}} \), which integrates to \( \sqrt{u} \). Substituting back, we get \( \sqrt{1+x^2} \). We apply the same logic for the \( y \) term, which also integrates to \( \sqrt{1+y^2} \). So, the final solution is \( \sqrt{1+x^2} + \sqrt{1+y^2} = c \). Substitution is an effective way to handle square root terms in integrals.
In simple words: Separate 'x' and 'y' terms. Use a substitution (like \( u = 1+x² \)) to make the integral simpler for both parts. After integrating, put the original variables back. The answer will be two square root terms added together, equaling a constant.

🎯 Exam Tip: For integrals of the form \( \int \frac{x}{\sqrt{a^2+x^2}} dx \) or \( \int \frac{x}{\sqrt{x^2-a^2}} dx \), a direct substitution like \( u=a^2+x^2 \) (or \( u=x^2-a^2 \)) is highly efficient.

Question 15.
\( (e^x + 1)y dy = (y + 1)e^x dx \)
Answer: The equation is \( (e^x + 1)y dy = (y + 1)e^x dx \). First, we separate the variables by moving \( (y+1) \) to the left side and \( (e^x+1) \) to the right side. This gives \( \frac{y}{y+1} dy = \frac{e^x}{e^x+1} dx \). We can rewrite the left-hand side fraction as \( 1 - \frac{1}{y+1} \). Now, we integrate both sides. The integral of \( 1 - \frac{1}{y+1} \) is \( y - \log|y+1| \). For the right side, notice that \( e^x \) is the derivative of \( e^x+1 \), so its integral is \( \log|e^x+1| \). Adding an integration constant \( c \), the solution is \( y - \log|y+1| = \log|e^x+1| + c \). This problem shows how to simplify fractions and recognize the \( \frac{f'(x)}{f(x)} \) integral form.
In simple words: Separate 'y' terms and 'x' terms. Rewrite the 'y' fraction to make it easier to integrate. Notice that the top of the 'x' fraction is the derivative of the bottom. Integrate both sides to get 'y' and 'log' terms.

🎯 Exam Tip: When faced with fractions like \( \frac{y}{y+1} \), try rewriting them as \( \frac{y+1-1}{y+1} = 1 - \frac{1}{y+1} \) to simplify integration.

Question 16.
(i) \( \log\left(\frac{d y}{d x}\right) = ax + by \)
(ii) \( \log\left(\frac{d y}{d x}\right) = 3x - 5y \)
Answer:
(i) We are given the equation \( \log\left(\frac{d y}{d x}\right) = ax + by \). First, we convert the logarithmic form into an exponential form: \( \frac{d y}{d x} = e^{ax+by} \). Using exponent rules, this can be written as \( \frac{d y}{d x} = e^{ax} e^{by} \). Next, we separate the variables, moving all \( y \) terms to the left side and all \( x \) terms to the right side. This gives us \( e^{-by} dy = e^{ax} dx \). Now, we integrate both sides. The integral of \( e^{-by} \) is \( \frac{e^{-by}}{-b} \), and the integral of \( e^{ax} \) is \( \frac{e^{ax}}{a} \). Adding an integration constant \( k \), we get \( \frac{e^{-by}}{-b} = \frac{e^{ax}}{a} + k \). Multiplying by \( -ab \) to clear the denominators and rearranging terms, we get the solution \( a e^{-by} + b e^{ax} = C \), where \( C \) is a new constant. This method converts logarithmic equations into a separable form.
(ii) We begin with the equation \( \log\left(\frac{d y}{d x}\right) = 3x - 5y \). Similar to the previous part, we convert this logarithmic equation into its exponential form: \( \frac{d y}{d x} = e^{3x-5y} \). Using exponent rules, this can be written as \( \frac{d y}{d x} = e^{3x} e^{-5y} \). Next, we separate the variables, bringing all \( y \) terms to the left side and all \( x \) terms to the right side. This gives us \( e^{5y} dy = e^{3x} dx \). Now, we integrate both sides. The integral of \( e^{5y} \) is \( \frac{e^{5y}}{5} \), and the integral of \( e^{3x} \) is \( \frac{e^{3x}}{3} \). Adding an integration constant \( c \), the solution is \( \frac{e^{5y}}{5} = \frac{e^{3x}}{3} + c \). This can be rearranged to \( \frac{e^{5y}}{5} - \frac{e^{3x}}{3} = c \). Converting log to exponent form helps in solving differential equations.
In simple words: For both parts, change the 'log' equation into an 'e' (exponential) equation. Separate 'y' terms and 'x' terms. Integrate both sides. For (i), multiply to clear fractions and rearrange. For (ii), rearrange to get 'e' terms and a constant.

🎯 Exam Tip: Exponential forms are often easier to separate and integrate. Remember the conversion: \( \log_b A = C \iff A = b^C \). For natural logarithm, \( \ln A = C \iff A = e^C \).

Question 17.
\( \frac { dy }{ dx } = \cos³ x \sin⁴ x + x\sqrt{2x+1} \)
Answer: We have the differential equation \( \frac { dy }{ dx } = \cos³ x \sin⁴ x + x\sqrt{2x+1} \). This equation is already in a form where variables are separated, so we just need to integrate both sides with respect to \( x \). The integral on the left side is simply \( y \). The integral on the right side involves two separate parts. For the first part, \( \int \cos³ x \sin⁴ x dx \), we rewrite \( \cos³ x \) as \( (1-\sin² x) \cos x \) and use the substitution \( t = \sin x \). This transforms it into \( \int (t⁴ - t⁶) dt \), which integrates to \( \frac{t⁵}{5} - \frac{t⁷}{7} \). Substituting \( t \) back, we get \( \frac{\sin⁵ x}{5} - \frac{\sin⁷ x}{7} \). For the second part, \( \int x\sqrt{2x+1} dx \), we use the substitution \( u = 2x+1 \). This integral simplifies and integrates to \( \frac{1}{10}(2x+1)^{5/2} - \frac{1}{6}(2x+1)^{3/2} \). Combining both parts and adding an integration constant \( C \), we get the full solution for \( y \). This problem requires mastering various integration techniques, including substitution and trigonometric identities.
In simple words: Separate the equation into two parts. For the first part, use 'sin x' as a temporary variable to make it easier to integrate. For the second part, use a substitution like \( u = 2x+1 \) to simplify the square root. Integrate both parts and add them together with a constant to get the final answer for \( y \).

🎯 Exam Tip: When integrating powers of sine and cosine, look for odd powers. If \( \cos^m x \sin^n x \) has \( m \) odd, substitute \( u = \sin x \). If \( n \) is odd, substitute \( u = \cos x \).

Question 18.
\( \sqrt{1+x^2+y^2+x^2 y^2}+x y \frac{d y}{d x} = 0 \)
Answer: The differential equation is \( \sqrt{1+x^2+y^2+x^2 y^2}+x y \frac{d y}{d x} = 0 \). First, we factor the terms under the square root to get \( \sqrt{(1+x²)(1+y²)} \), which simplifies to \( \sqrt{1+x²}\sqrt{1+y²} \). So the equation becomes \( \sqrt{1+x²}\sqrt{1+y²} + xy \frac{d y}{d x} = 0 \). We rearrange and separate variables: \( \frac{y}{\sqrt{1+y²}} dy = -\frac{\sqrt{1+x²}}{x} dx \). Now, we integrate both sides. The left side, \( \int \frac{y}{\sqrt{1+y²}} dy \), integrates to \( \sqrt{1+y^2} \) using a substitution similar to Question 14. For the right side, \( -\int \frac{\sqrt{1+x²}}{x} dx \), we use the substitution \( u = \sqrt{1+x²} \). This transforms the integral into \( -\int \left(1+\frac{1}{u²-1}\right) du \), which integrates to \( -\left(u + \frac{1}{2}\log\left|\frac{u-1}{u+1}\right|\right) \). Substituting back for \( u \) and combining with the left side, we get the solution \( \sqrt{1+x²} + \sqrt{1+y²} + \frac{1}{2}\log\left|\frac{\sqrt{1+x²}-1}{\sqrt{1+x²}+1}\right| = C \). This complex problem showcases advanced integration techniques using substitutions.
In simple words: Factor the square root term. Rearrange to separate 'y' terms with 'dy' and 'x' terms with 'dx'. Integrate the 'y' part (it's similar to Q14). For the 'x' part, use a substitution like \( u = \sqrt{1+x²} \), which will make the integral simpler to solve using partial fractions. Put everything together to get the final answer.

🎯 Exam Tip: Always look for factoring opportunities to simplify complex terms, especially those under square roots, as it often makes the equation separable.

Question 19.
\( \frac{d y}{d x}=\frac{e^x\left(\sin ^2 x+\sin 2 x\right)}{y(2 \log y+1)} \)
Answer: We are given the differential equation \( \frac{d y}{d x}=\frac{e^x\left(\sin ^2 x+\sin 2 x\right)}{y(2 \log y+1)} \). First, we separate the variables by moving all \( y \) terms to the left side and all \( x \) terms to the right. This gives us \( y(2 \log y+1) dy = e^x(\sin² x + \sin 2x) dx \). Now, we integrate both sides. For the left side, \( \int y(2 \log y+1) dy \), we can expand it to \( \int (2y \log y + y) dy \). Using integration by parts for \( \int 2y \log y dy \) (with \( U = \log y \) and \( dV = 2y dy \)), it evaluates to \( y² \log y - \frac{y²}{2} \). Adding the integral of \( y \) (which is \( \frac{y²}{2} \)), the left side simplifies to \( y² \log y \). For the right side, \( \int e^x(\sin² x + \sin 2x) dx \), we notice it is of the form \( \int e^x(f(x) + f'(x)) dx \). Here, if \( f(x) = \sin² x \), then \( f'(x) = 2 \sin x \cos x = \sin 2x \). So, the integral is \( e^x \sin² x \). Combining both sides with an integration constant \( c \), the solution is \( y² \log y = e^x \sin² x + c \). This problem highlights the use of integration by parts and the special \( \int e^x(f(x) + f'(x)) dx \) formula.
In simple words: Separate 'y' terms and 'x' terms. Integrate the 'y' side using a special rule for parts. For the 'x' side, notice it's a special form where \( e^x \) is multiplied by a function plus its derivative. Integrate both sides and add a constant to get the answer.

🎯 Exam Tip: Master the integration by parts formula \( \int u dv = uv - \int v du \) and recognize the special form \( \int e^x(f(x) + f'(x)) dx = e^x f(x) \) as they are commonly used in differential equations.

Question 20.
\( \sqrt{1+x^2} dy+\sqrt{1+y^2} dx = 0 \)
Answer: The differential equation is \( \sqrt{1+x^2} dy+\sqrt{1+y^2} dx = 0 \). First, we separate the variables by dividing the entire equation by \( \sqrt{1+x^2}\sqrt{1+y^2} \). This results in \( \frac{dy}{\sqrt{1+y^2}} + \frac{dx}{\sqrt{1+x^2}} = 0 \). Now, we integrate both sides. The integral of \( \frac{dy}{\sqrt{1+y^2}} \) is a standard form, \( \log|y+\sqrt{1+y^2}| \). Similarly, for the \( x \) term, it is \( \log|x+\sqrt{1+x^2}| \). We use \( \log C \) as the integration constant. So, we have \( \log|y+\sqrt{1+y^2}| + \log|x+\sqrt{1+x^2}| = \log C \). Using logarithm properties, we combine the terms: \( \log((y+\sqrt{1+y^2})(x+\sqrt{1+x^2})) = \log C \). Removing the logarithms from both sides gives us the final solution: \( (y+\sqrt{1+y^2})(x+\sqrt{1+x^2}) = C \). This problem relies on recognizing standard integral forms involving square roots.
In simple words: Separate 'y' terms with 'dy' and 'x' terms with 'dx'. Integrate both sides using a standard formula that gives 'log' terms with square roots. Combine the 'log' terms and remove 'log' to get the final answer.

🎯 Exam Tip: Recognize and memorize standard integral formulas, such as \( \int \frac{dx}{\sqrt{a^2+x^2}} = \log|x+\sqrt{a^2+x^2}| \), to quickly solve such problems.

Question 21.
\( \frac{d y}{d x}-x \sin ^2 x=\frac{1}{x \log x} \)
Answer: We are given the equation \( \frac{d y}{d x}-x \sin ^2 x=\frac{1}{x \log x} \). First, we rearrange it to solve for \( \frac{d y}{d x} \): \( \frac{d y}{d x} = x \sin ^2 x + \frac{1}{x \log x} \). This equation is already in a form where we can integrate both sides with respect to \( x \). The integral of \( \frac{d y}{d x} \) is \( y \). For the right side, we integrate each term separately. For \( \int x \sin ^2 x dx \), we use the identity \( \sin² x = \frac{1-\cos 2x}{2} \) and then apply integration by parts to \( \int x \cos 2x dx \). This part gives \( \frac{x²}{4} - \frac{x}{4}\sin 2x - \frac{1}{8}\cos 2x \). For the second part, \( \int \frac{1}{x \log x} dx \), we use the substitution \( t = \log x \), which simplifies the integral to \( \int \frac{1}{t} dt = \log|t| = \log|\log x| \). Combining these parts and adding an integration constant \( c \), the full solution for \( y \) is \( y = \frac{x²}{4} - \frac{x}{4}\sin 2x - \frac{1}{8}\cos 2x + \log|\log x| + c \). This problem demonstrates the need for multiple integration methods within a single solution.
In simple words: Rearrange the equation to get \( \frac{dy}{dx} \) alone. Then, integrate both sides. For the \( x \sin² x \) part, use a trig rule to change \( \sin² x \) and then integrate by parts. For the \( \frac{1}{x \log x} \) part, use a substitution for \( \log x \). Add all integrated parts and a constant to get the final answer for \( y \).

🎯 Exam Tip: When integrating \( x \sin^2 x \), transform \( \sin^2 x \) using the double-angle formula \( \sin^2 x = \frac{1-\cos 2x}{2} \) before applying integration by parts.

Find the Particular Solution of the Following Differential Equations:

Question 22.
(i) \( \cos y dy + \cos x \sin y dx = 0, y(\frac { π }{ 2 }) = \frac { π }{ 2 } \)
(ii) \( (1 - x²)\frac { dy }{ dx } - xy = x; \text{given } y = 1 \text{ when } x = 0. \)
(iii) \( (x² - yx²)dy + (y² + x²y²)dx = 0, \text{given that } y = 1, \text{ when } x = 1. \)
(iv) \( \frac { dy }{ dx } = 1 + x² + y² + x²y², \text{given that } y = 1 \text{ when } x = 0. \)
(v) \( (1 + e²x) dy + (1 + y²)exdx = 0, \text{given that } y = 1 \text{ when } x = 0. \)
(vi) \( \frac { dy }{ dx } = 1 + x + y + xy, \text{given that } y = 1 \text{ when } x = 0. \)
(vii) \( \frac { dy }{ dx } = \frac{x(2 \log |x|+1)}{\sin y+y \cos y}, \text{given that } y = \frac { π }{ 2 }, \text{ when } x = 1. \)
Answer:
(i) We are given the differential equation \( \cos y dy + \cos x \sin y dx = 0 \) along with an initial condition \( y(\frac { π }{ 2 }) = \frac { π }{ 2 } \). First, we separate the variables by dividing by \( \sin y \), which gives \( \frac{\cos y}{\sin y} dy + \cos x dx = 0 \). This can be written as \( \cot y dy + \cos x dx = 0 \). Now, we integrate both sides. The integral of \( \cot y \) is \( \log|\sin y| \), and the integral of \( \cos x \) is \( \sin x \). So, the general solution is \( \log|\sin y| + \sin x = c \). To find the *particular* solution, we use the given condition \( x = \frac{π}{2} \) and \( y = \frac{π}{2} \). Substituting these values into the general solution: \( \log|\sin(\frac{π}{2})| + \sin(\frac{π}{2}) = c \). Since \( \sin(\frac{π}{2}) = 1 \) and \( \log|1| = 0 \), we find that \( 0 + 1 = c \), so \( c = 1 \). Therefore, the particular solution is \( \log|\sin y| + \sin x = 1 \). Initial conditions help us find the exact solution from a family of possible solutions.
(ii) We have the equation \( (1 - x²)\frac { dy }{ dx } - xy = x \) with the condition \( y = 1 \) when \( x = 0 \). First, we rearrange the equation to \( (1 - x²)\frac { dy }{ dx } = x(1 + y) \). Then, we separate the variables, putting \( y \) terms with \( dy \) and \( x \) terms with \( dx \). This gives \( \frac{dy}{1+y} = \frac{x}{1-x²} dx \). Next, we integrate both sides. The integral of \( \frac{1}{1+y} \) is \( \log|1+y| \). For the right side, we adjust the numerator to be the derivative of the denominator (by multiplying by -2 and \( -\frac{1}{2} \)), making its integral \( -\frac{1}{2} \log|1-x²| \). So, the general solution is \( \log|1+y| = -\frac{1}{2} \log|1-x²| + c \). Now, we use the condition \( x=0, y=1 \) to find \( c \). Substituting these values, we get \( \log|1+1| = -\frac{1}{2} \log|1-0²| + c \), which simplifies to \( \log 2 = 0 + c \), so \( c = \log 2 \). Plugging this back, the particular solution is \( \log|1+y| = -\frac{1}{2} \log|1-x²| + \log 2 \). We can rearrange this using logarithm properties to \( (1+y)\sqrt{1-x²} = 2 \). Initial values are important for unique solutions.
(iii) We are given the equation \( (x² - yx²)dy + (y² + x²y²)dx = 0 \) with the condition \( y = 1 \) when \( x = 1 \). First, we factor out common terms: \( x²(1 - y)dy + y²(1 + x²)dx = 0 \). To separate variables, we divide the entire equation by \( x²y² \), resulting in \( \frac{1-y}{y^2} dy + \frac{1+x²}{x^2} dx = 0 \). We can split these fractions into simpler terms: \( \left(\frac{1}{y^2}-\frac{1}{y}\right) dy + \left(\frac{1}{x^2}+1\right) dx = 0 \). Now, we integrate each term. This gives us \( -\frac{1}{y} - \log|y| - \frac{1}{x} + x = c \), which is the general solution. To find the particular solution, we use the condition \( x=1, y=1 \). Substituting these values: \( -\frac{1}{1} - \log|1| - \frac{1}{1} + 1 = c \). This simplifies to \( -1 - 0 - 1 + 1 = c \), so \( c = -1 \). Therefore, the particular solution is \( -\frac{1}{y} - \log|y| - \frac{1}{x} + x = -1 \). Using initial conditions helps determine the specific behavior of the system.
In simple words: For each part, first separate 'x' and 'y' terms. Integrate to find the general solution with a constant 'c'. Then, use the given starting values (initial conditions) for 'x' and 'y' to calculate the exact value of 'c'. Finally, substitute this value of 'c' back into the general solution to get the particular solution. Remember to simplify the final answer using logarithm properties.

🎯 Exam Tip: The key difference between general and particular solutions is the constant of integration. A general solution contains an arbitrary constant (C), while a particular solution determines the specific value of C using initial conditions.

Question 22. Find the particular solution of the following differential equations :
(iv) \( \frac { dy }{ dx } = 1 + x² + y² + x²y², \) given that \( y = 1 \) when \( x = 0. \)
Answer:
Given the differential equation: \( \frac { dy }{ dx } = 1 + x² + y² + x²y² \)
This can be rewritten as: \( \frac { dy }{ dx } = (1 + x²)(1 + y²) \)
Now, we separate the variables:
\( \frac { dy }{ 1+y² } = (1 + x²) dx \)
Next, integrate both sides:
\( \int \frac { dy }{ 1+y² } = \int (1+x²) dx \)
\( \implies tan^{-1} y = x + \frac { x³ }{ 3 } + c \)
We use the given condition that \( y = 1 \) when \( x = 0 \) to find the value of \( c \):
\( tan^{-1} 1 = 0 + \frac { 0³ }{ 3 } + c \)
\( \implies \frac { \pi }{ 4 } = c \)
Substitute the value of \( c \) back into the general solution:
\( tan^{-1} y = x + \frac { x³ }{ 3 } + \frac { \pi }{ 4 } \)
Finally, solve for \( y \):
\( \implies y = tan \left( x + \frac { x³ }{ 3 } + \frac { \pi }{ 4 } \right) \)
This is the particular solution for the given differential equation.
In simple words: First, we rearrange the equation so all 'y' terms are on one side and 'x' terms are on the other. Then, we integrate both sides. Using the given starting values for x and y, we find the constant 'c'. We put this 'c' back into our integrated equation to get the final answer.

🎯 Exam Tip: Remember to simplify the right side of the differential equation by factoring (like \( (1+x²)(1+y²) \)) before separating variables. This makes integration much easier.

Question 22. Find the particular solution of the following differential equations :
(v) \( (1 + e^{2x}) dy + (1 + y²)e^x dx = 0, \) given that \( y = 1 \) when \( x = 0. \)
Answer:
Given the differential equation: \( (1 + e^{2x}) dy + (1 + y²)e^x dx = 0 \)
First, we separate the variables:
\( \frac { dy }{ 1+y² } + \frac { e^x dx }{ 1+e^{2x} } = 0 \)
Now, we integrate both sides:
\( \int \frac { dy }{ 1+y² } + \int \frac { e^x dx }{ 1+e^{2x} } = c \)
For the second integral, let \( t = e^x \). Then \( dt = e^x dx \). The integral becomes \( \int \frac { dt }{ 1+t² } \).
So, integrating gives:
\( \implies tan^{-1} y + tan^{-1} t = c \)
Substitute back \( t = e^x \):
\( \implies tan^{-1} y + tan^{-1} (e^x) = c \)
Next, use the given condition that \( y = 1 \) when \( x = 0 \) to find \( c \):
\( tan^{-1} 1 + tan^{-1} (e^0) = c \)
\( \implies \frac { \pi }{ 4 } + \frac { \pi }{ 4 } = c \)
\( \implies c = \frac { \pi }{ 2 } \)
Substitute \( c \) back into the general solution:
\( tan^{-1} y + tan^{-1} (e^x) = \frac { \pi }{ 2 } \)
Using the identity \( tan^{-1} A + tan^{-1} B = tan^{-1} \left( \frac { A+B }{ 1-AB } \right) \):
\( \implies tan^{-1} \left( \frac { y+e^x }{ 1-ye^x } \right) = \frac { \pi }{ 2 } \)
For \( tan^{-1} Z = \frac{\pi}{2} \), Z must approach infinity, which means the denominator must be zero.
\( \implies 1-ye^x = 0 \)
\( \implies ye^x = 1 \)
\( \implies y = e^{-x} \)
This is the particular solution. The tangent of \( \frac{\pi}{2} \) is undefined, so the expression inside \( tan^{-1} \) must have a denominator of zero.
In simple words: First, group all 'y' terms with 'dy' and 'x' terms with 'dx'. Then, integrate both sides. To find the special constant, put the given values of x and y into your answer. If the tangent of an angle is \( \frac{\pi}{2} \), it means the bottom part of the fraction inside the tangent must be zero.

🎯 Exam Tip: When dealing with \( tan^{-1} \frac{\pi}{2} \), remember that \( tan(\frac{\pi}{2}) \) is undefined. This implies that the argument of the tangent function must be infinitely large, meaning its denominator must be zero.

Question 22. Find the particular solution of the following differential equations :
(vi) \( \frac { dy }{ dx } = 1 + x + y + xy, \) given that \( y = 0 \) when \( x = 1. \)
Answer:
Given the differential equation: \( \frac { dy }{ dx } = 1 + x + y + xy \)
Factor the right side:
\( \frac { dy }{ dx } = (1 + x) + y(1 + x) \)
\( \frac { dy }{ dx } = (1 + x)(1 + y) \)
Now, separate the variables:
\( \frac { dy }{ 1+y } = (1 + x)dx \)
Integrate both sides:
\( \int \frac { dy }{ 1+y } = \int (1+x)dx \)
\( \implies log | 1 + y | = x + \frac { x² }{ 2 } + c \)
Use the given condition that \( y = 0 \) when \( x = 1 \) to find \( c \):
\( log | 1 + 0 | = 1 + \frac { 1² }{ 2 } + c \)
\( log 1 = 1 + \frac { 1 }{ 2 } + c \)
\( 0 = \frac { 3 }{ 2 } + c \)
\( \implies c = -\frac { 3 }{ 2 } \)
Substitute \( c \) back into the general solution:
\( log | 1 + y | = x + \frac { x² }{ 2 } - \frac { 3 }{ 2 } \)
This is the particular solution.
In simple words: The first step is to rearrange the equation to put all 'y' terms with 'dy' and 'x' terms with 'dx'. Then, you integrate both sides. Use the starting values of x and y given in the problem to find the special constant 'c'. Finally, substitute 'c' back into your integrated equation to get the full answer.

🎯 Exam Tip: Always look for ways to factorize expressions on the right-hand side, as this often simplifies the separation of variables, which is key for solving these types of differential equations.

Question 22. Find the particular solution of the following differential equations :
(vii) \( \frac { dy }{ dx } = \frac{x(2 \log |x|+1)}{\sin y+y \cos y}, \) given that \( y = \frac { \pi }{ 2 }, \) when \( x = 1. \)
Answer:
Given the differential equation: \( \frac { dy }{ dx } = \frac{x(2 \log |x|+1)}{\sin y+y \cos y} \)
First, separate the variables:
\( (\sin y + y \cos y) dy = x (2 \log | x | + 1) dx \)
Now, integrate both sides:
\( \int (\sin y + y \cos y) dy = \int x (2 \log | x | + 1) dx \)
Break down the left side integral: \( \int \sin y dy + \int y \cos y dy \)
We know \( \int \sin y dy = -\cos y \).
For \( \int y \cos y dy \), use integration by parts \( (\int u dv = uv - \int v du) \). Let \( u = y \), \( dv = \cos y dy \). Then \( du = dy \), \( v = \sin y \).
So, \( \int y \cos y dy = y \sin y - \int \sin y dy = y \sin y - (-\cos y) = y \sin y + \cos y \).
Thus, the left side integral is \( -\cos y + y \sin y + \cos y = y \sin y \).
Break down the right side integral: \( 2 \int x \log | x | dx + \int x dx \)
We know \( \int x dx = \frac{x^2}{2} \).
For \( \int x \log | x | dx \), use integration by parts. Let \( u = \log |x| \), \( dv = x dx \). Then \( du = \frac{1}{x} dx \), \( v = \frac{x^2}{2} \).
So, \( \int x \log | x | dx = \log |x| \frac{x^2}{2} - \int \frac{x^2}{2} \frac{1}{x} dx = \frac{x^2}{2} \log |x| - \int \frac{x}{2} dx = \frac{x^2}{2} \log |x| - \frac{x^2}{4} \).
Thus, the right side integral is \( 2 \left( \frac{x^2}{2} \log |x| - \frac{x^2}{4} \right) + \frac{x^2}{2} = x^2 \log |x| - \frac{x^2}{2} + \frac{x^2}{2} = x^2 \log |x| \).
Combining both sides, we get the general solution:
\( y \sin y = x² \log | x | + c \)
Now, use the given condition that \( y = \frac { \pi }{ 2 } \) when \( x = 1 \) to find \( c \):
\( \frac { \pi }{ 2 } \sin \frac { \pi }{ 2 } = 1² \log | 1 | + c \)
\( \frac { \pi }{ 2 } (1) = 1(0) + c \)
\( \implies c = \frac { \pi }{ 2 } \)
Substitute \( c \) back into the general solution:
\( y \sin y = x² \log | x | + \frac { \pi }{ 2 } \)
This is the particular solution.
In simple words: This problem involves separating the 'y' and 'x' parts, then integrating both sides. The trickiest part is using "integration by parts" for both the 'y' and 'x' side. After integrating, you use the given x and y values to find the constant 'c' and put it back into your answer.

🎯 Exam Tip: For problems involving \( y \cos y \) or \( x \log x \) terms, remember to apply integration by parts carefully. Recognizing the derivative of \( y \sin y \) (which is \( \sin y + y \cos y \)) can also simplify the left-hand side integral directly.

Question 23. \( \frac { dy }{ dx } = (2x + 3y – 4)² \)
Answer:
Given the differential equation: \( \frac { dy }{ dx } = (2x + 3y – 4)² \)
This type of equation can be solved by substitution. Let \( t = 2x + 3y - 4 \).
Differentiate \( t \) with respect to \( x \):
\( \frac { dt }{ dx } = 2 + 3 \frac { dy }{ dx } - 0 \)
\( \implies \frac { dy }{ dx } = \frac { 1 }{ 3 } \left( \frac { dt }{ dx } - 2 \right) \)
Substitute \( \frac { dy }{ dx } \) and \( t \) into the original equation:
\( \frac { 1 }{ 3 } \left( \frac { dt }{ dx } - 2 \right) = t² \)
\( \implies \frac { dt }{ dx } - 2 = 3t² \)
\( \implies \frac { dt }{ dx } = 3t² + 2 \)
Now, separate the variables:
\( \frac { dt }{ 3t² + 2 } = dx \)
Integrate both sides:
\( \int \frac { dt }{ 3t² + 2 } = \int dx \)
Factor out 3 from the denominator on the left side:
\( \frac { 1 }{ 3 } \int \frac { dt }{ t² + \frac { 2 }{ 3 } } = x + c \)
This integral is of the form \( \int \frac { dz }{ z² + a² } = \frac { 1 }{ a } tan^{-1} \left( \frac { z }{ a } \right) \), where \( a = \sqrt { \frac { 2 }{ 3 } } \).
\( \implies \frac { 1 }{ 3 } \cdot \frac { 1 }{ \sqrt { \frac { 2 }{ 3 } } } tan^{-1} \left( \frac { t }{ \sqrt { \frac { 2 }{ 3 } } } \right) = x + c \)
\( \implies \frac { 1 }{ 3 } \sqrt { \frac { 3 }{ 2 } } tan^{-1} \left( t \sqrt { \frac { 3 }{ 2 } } \right) = x + c \)
\( \implies \frac { 1 }{ \sqrt { 3 } \sqrt { 2 } } tan^{-1} \left( \sqrt { \frac { 3 }{ 2 } } t \right) = x + c \)
\( \implies \frac { 1 }{ \sqrt { 6 } } tan^{-1} \left( \sqrt { \frac { 3 }{ 2 } } t \right) = x + c \)
Finally, substitute back \( t = 2x + 3y - 4 \):
\( \frac { 1 }{ \sqrt { 6 } } tan^{-1} \left( \sqrt { \frac { 3 }{ 2 } } (2x+3y-4) \right) = x + c \)
This is the required general solution.
In simple words: When you see a differential equation where \( dy/dx \) equals a square of a linear expression like \( (ax+by+c)^2 \), use substitution to simplify it. Replace the linear part with a new variable, then find its derivative. This will turn the original equation into one where you can separate the variables and integrate.

🎯 Exam Tip: Recognize when substitution \( t = ax+by+c \) is useful. After substitution, remember the integral formula \( \int \frac{1}{z^2+a^2} dz = \frac{1}{a} \tan^{-1} (\frac{z}{a}) \). Don't forget to substitute back the original variables at the end.

Question 24. \( (x + y + 1)\frac { dy }{ dx } = 1 \)
Answer:
Given the differential equation: \( (x + y + 1)\frac { dy }{ dx } = 1 \)
Let \( t = x + y + 1 \). This type of substitution simplifies the equation.
Differentiate \( t \) with respect to \( x \):
\( \frac { dt }{ dx } = 1 + \frac { dy }{ dx } \)
\( \implies \frac { dy }{ dx } = \frac { dt }{ dx } - 1 \)
Substitute \( t \) and \( \frac { dy }{ dx } \) into the original equation:
\( t \left( \frac { dt }{ dx } - 1 \right) = 1 \)
\( \implies \frac { dt }{ dx } - 1 = \frac { 1 }{ t } \)
\( \implies \frac { dt }{ dx } = 1 + \frac { 1 }{ t } \)
Combine terms on the right side:
\( \frac { dt }{ dx } = \frac { t+1 }{ t } \)
Now, separate the variables:
\( \frac { t }{ t+1 } dt = dx \)
Integrate both sides:
\( \int \frac { t }{ t+1 } dt = \int dx \)
To integrate the left side, rewrite the numerator:
\( \int \frac { t+1-1 }{ t+1 } dt = x + c \)
\( \int \left( 1 - \frac { 1 }{ t+1 } \right) dt = x + c \)
\( \implies t - log | t+1 | = x + c \)
Finally, substitute back \( t = x + y + 1 \):
\( (x + y + 1) - log | (x + y + 1)+1 | = x + c \)
\( x + y + 1 - log | x + y + 2 | = x + c \)
Simplify the equation:
\( y + 1 - log | x + y + 2 | = c \)
Move the constant to the right side, defining a new constant \( c' = c - 1 \):
\( y - log | x + y + 2 | = c' \)
This is the required general solution.
In simple words: When the differential equation has \( (x+y) \) or \( (x+y+c) \) in it, try replacing that part with a new variable. This helps turn a complicated equation into one where you can separate the 'x' and 't' terms. After integrating, just put the original expression back in place of 't'.

🎯 Exam Tip: Always look for terms like \( (x+y) \) or \( (x+y+k) \) as a signal for using the substitution \( t=x+y \) or \( t=x+y+k \). This simplifies the equation to a separable form, allowing integration.

Question 25. \( (x + y)² \frac { dy }{ dx } = 1 \)
Answer:
Given the differential equation: \( (x + y)² \frac { dy }{ dx } = 1 \)
Let \( t = x + y \). This is a common substitution for such equations.
Differentiate \( t \) with respect to \( x \):
\( \frac { dt }{ dx } = 1 + \frac { dy }{ dx } \)
\( \implies \frac { dy }{ dx } = \frac { dt }{ dx } - 1 \)
Substitute \( t \) and \( \frac { dy }{ dx } \) into the original equation:
\( t² \left( \frac { dt }{ dx } - 1 \right) = 1 \)
\( \implies \frac { dt }{ dx } - 1 = \frac { 1 }{ t² } \)
\( \implies \frac { dt }{ dx } = 1 + \frac { 1 }{ t² } \)
Combine terms on the right side:
\( \frac { dt }{ dx } = \frac { t² + 1 }{ t² } \)
Now, separate the variables:
\( \frac { t² }{ t² + 1 } dt = dx \)
Integrate both sides:
\( \int \frac { t² }{ t² + 1 } dt = \int dx \)
To integrate the left side, rewrite the numerator:
\( \int \frac { t² + 1 - 1 }{ t² + 1 } dt = x + c \)
\( \int \left( 1 - \frac { 1 }{ t² + 1 } \right) dt = x + c \)
\( \implies t - tan^{-1} t = x + c \)
Finally, substitute back \( t = x + y \):
\( (x + y) - tan^{-1} (x + y) = x + c \)
Simplify the equation:
\( y - tan^{-1} (x + y) = c \)
This is the required general solution.
In simple words: For equations where \( dy/dx \) depends on \( (x+y) \), it's a good idea to substitute \( t = x+y \). This helps to transform the equation into a simpler form where you can group 't' terms with 'dt' and 'x' terms with 'dx'. After integrating, remember to replace 't' with \( (x+y) \) to get the final solution.

🎯 Exam Tip: When using the substitution \( t = x+y \), remember that \( \frac{dt}{dx} = 1 + \frac{dy}{dx} \). Also, be proficient with integrating fractions like \( \frac{t^2}{t^2+1} \) by adding and subtracting 1 in the numerator.

Question 26. \( \frac { dy }{ dx } = \cos(x + y) \)
Answer:
Given the differential equation: \( \frac { dy }{ dx } = \cos(x + y) \)
Let \( t = x + y \). This substitution helps to simplify the argument of the cosine function.
Differentiate \( t \) with respect to \( x \):
\( \frac { dt }{ dx } = 1 + \frac { dy }{ dx } \)
\( \implies \frac { dy }{ dx } = \frac { dt }{ dx } - 1 \)
Substitute \( t \) and \( \frac { dy }{ dx } \) into the original equation:
\( \frac { dt }{ dx } - 1 = \cos t \)
\( \implies \frac { dt }{ dx } = 1 + \cos t \)
Now, separate the variables:
\( \frac { dt }{ 1 + \cos t } = dx \)
Integrate both sides:
\( \int \frac { dt }{ 1 + \cos t } = \int dx \)
Use the half-angle identity \( 1 + \cos t = 2 \cos² \frac { t }{ 2 } \):
\( \int \frac { dt }{ 2 \cos² \frac { t }{ 2 } } = x + c \)
\( \int \frac { 1 }{ 2 } \sec² \frac { t }{ 2 } dt = x + c \)
Integrate the left side:
\( \frac { 1 }{ 2 } \cdot \frac { \tan \frac { t }{ 2 } }{ \frac { 1 }{ 2 } } = x + c \)
\( \implies \tan \frac { t }{ 2 } = x + c \)
Finally, substitute back \( t = x + y \):
\( \tan \left( \frac { x+y }{ 2 } \right) = x + c \)
This is the required general solution.
In simple words: When you see \( (x+y) \) inside a trigonometric function in a differential equation, replacing \( (x+y) \) with a new variable 't' can simplify things. Then, you can separate the 't' terms from the 'x' terms and integrate. Remember to use trigonometric identities like \( 1+\cos t = 2\cos^2(t/2) \) to make integration easier.

🎯 Exam Tip: The substitution \( t=x+y \) is effective for equations involving \( f(x+y) \). Remember the trigonometric identity \( 1+\cos \theta = 2\cos^2(\theta/2) \) for simplifying integrals with \( 1+\cos t \) in the denominator.

Question 27. \( \frac { dy }{ dx } = \tan²(x + y) \)
Answer:
Given the differential equation: \( \frac { dy }{ dx } = \tan²(x + y) \)
Let \( t = x + y \). This substitution will simplify the equation.
Differentiate \( t \) with respect to \( x \):
\( \frac { dt }{ dx } = 1 + \frac { dy }{ dx } \)
\( \implies \frac { dy }{ dx } = \frac { dt }{ dx } - 1 \)
Substitute \( t \) and \( \frac { dy }{ dx } \) into the original equation:
\( \frac { dt }{ dx } - 1 = \tan² t \)
\( \implies \frac { dt }{ dx } = 1 + \tan² t \)
Using the trigonometric identity \( 1 + \tan² t = \sec² t \):
\( \frac { dt }{ dx } = \sec² t \)
Now, separate the variables:
\( \frac { dt }{ \sec² t } = dx \)
\( \implies \cos² t dt = dx \)
Integrate both sides:
\( \int \cos² t dt = \int dx \)
Use the double-angle identity \( \cos² t = \frac { 1 + \cos 2t }{ 2 } \):
\( \int \frac { 1 + \cos 2t }{ 2 } dt = x + c \)
\( \frac { 1 }{ 2 } \int (1 + \cos 2t) dt = x + c \)
\( \frac { 1 }{ 2 } \left( t + \frac { \sin 2t }{ 2 } \right) = x + c \)
Multiply by 2:
\( t + \frac { \sin 2t }{ 2 } = 2x + 2c \)
Finally, substitute back \( t = x + y \):
\( (x + y) + \frac { \sin 2(x+y) }{ 2 } = 2x + 2c \)
Rearrange the terms:
\( y - x + \frac { \sin 2(x+y) }{ 2 } = 2c \)
Multiply by 2 and define a new constant \( c' = 4c \):
\( 2(y-x) + \sin 2(x+y) = 4c \)
\( \implies 2(y-x) + \sin 2(x+y) = c' \)
This is the required general solution.
In simple words: This problem becomes easier if you replace \( (x+y) \) with a new variable 't'. Then, you can use the identity \( 1+\tan^2 t = \sec^2 t \) to simplify the equation. After separating 't' and 'x' terms, you integrate using the identity \( \cos^2 t = \frac{1+\cos 2t}{2} \). Remember to put \( (x+y) \) back in place of 't' at the end.

🎯 Exam Tip: For equations with \( \tan^2(x+y) \), the substitution \( t=x+y \) transforms it to a standard integrable form. Remember the identity \( \cos^2 A = \frac{1+\cos 2A}{2} \) which is frequently used in integration to handle even powers of cosine.

Question 28. \( \cos²(x – 2y) = 1 – 2\frac { dy }{ dx } \)
Answer:
Given the differential equation: \( \cos²(x – 2y) = 1 – 2\frac { dy }{ dx } \)
Let \( t = x - 2y \). This substitution helps simplify the argument of the cosine function.
Differentiate \( t \) with respect to \( x \):
\( \frac { dt }{ dx } = 1 - 2 \frac { dy }{ dx } \)
Notice that the right side of the original equation is exactly \( \frac{dt}{dx} \).
Substitute \( t \) and \( \frac { dt }{ dx } \) into the original equation:
\( \cos² t = \frac { dt }{ dx } \)
Now, separate the variables:
\( \frac { dt }{ \cos² t } = dx \)
\( \implies \sec² t dt = dx \)
Integrate both sides:
\( \int \sec² t dt = \int dx \)
\( \implies \tan t = x + c \)
Finally, substitute back \( t = x - 2y \):
\( \tan (x - 2y) = x + c \)
This is the required general solution.
In simple words: When you see \( (x-2y) \) inside a trigonometric function and the derivative \( 1-2(dy/dx) \) elsewhere, replacing \( (x-2y) \) with 't' will simplify the equation perfectly. This makes it easy to separate the 't' terms and 'x' terms and then integrate. Remember that the integral of \( \sec^2 t \) is \( \tan t \).

🎯 Exam Tip: Look for the derivative of your chosen substitution appearing directly in the equation. In this case, \( \frac{d}{dx}(x-2y) = 1 - 2\frac{dy}{dx} \), which matches the right side, simplifying the process greatly.

Question 29. \( \frac { dy }{ dx } = \frac{x+y+1}{2 x+2y+3} \)
Answer:
Given the differential equation: \( \frac { dy }{ dx } = \frac{x+y+1}{2 x+2y+3} \)
Notice that the expression \( x+y \) appears in both the numerator and denominator. We can rewrite the denominator as \( 2(x+y)+3 \).
Let \( t = x + y \). This is a helpful substitution for such equations.
Differentiate \( t \) with respect to \( x \):
\( \frac { dt }{ dx } = 1 + \frac { dy }{ dx } \)
\( \implies \frac { dy }{ dx } = \frac { dt }{ dx } - 1 \)
Substitute \( t \) and \( \frac { dy }{ dx } \) into the original equation:
\( \frac { dt }{ dx } - 1 = \frac { t+1 }{ 2t+3 } \)
\( \implies \frac { dt }{ dx } = 1 + \frac { t+1 }{ 2t+3 } \)
Combine terms on the right side by finding a common denominator:
\( \frac { dt }{ dx } = \frac { (2t+3) + (t+1) }{ 2t+3 } \)
\( \implies \frac { dt }{ dx } = \frac { 3t+4 }{ 2t+3 } \)
Now, separate the variables:
\( \frac { 2t+3 }{ 3t+4 } dt = dx \)
Integrate both sides:
\( \int \frac { 2t+3 }{ 3t+4 } dt = \int dx \)
To integrate the left side, adjust the numerator to match the denominator:
\( \int \frac { \frac { 2 }{ 3 } (3t+4) - \frac { 8 }{ 3 } + 3 }{ 3t+4 } dt = x + c \)
\( \int \frac { \frac { 2 }{ 3 } (3t+4) + \frac { 1 }{ 3 } }{ 3t+4 } dt = x + c \)
\( \int \left( \frac { 2 }{ 3 } + \frac { \frac { 1 }{ 3 } }{ 3t+4 } \right) dt = x + c \)
\( \implies \int \frac { 2 }{ 3 } dt + \frac { 1 }{ 3 } \int \frac { 1 }{ 3t+4 } dt = x + c \)
\( \implies \frac { 2 }{ 3 } t + \frac { 1 }{ 3 } \cdot \frac { \log | 3t+4 | }{ 3 } = x + c \)
\( \implies \frac { 2 }{ 3 } t + \frac { 1 }{ 9 } \log | 3t+4 | = x + c \)
Multiply the entire equation by 9 to clear denominators:
\( 6t + \log | 3t+4 | = 9x + 9c \)
Finally, substitute back \( t = x + y \):
\( 6(x + y) + \log | 3(x+y)+4 | = 9x + 9c \)
\( 6x + 6y + \log | 3x+3y+4 | = 9x + 9c \)
Rearrange the terms and define a new constant \( c' = 9c \):
\( 6y - 3x + \log | 3x+3y+4 | = c' \)
This is the required general solution.
In simple words: When the equation has \( (x+y) \) terms in both the top and bottom, substitute \( t = x+y \) to simplify. After substituting and rearranging, you'll get an integral where you might need to adjust the numerator to match the denominator, like turning \( \frac{2t+3}{3t+4} \) into \( \frac{2}{3} + \text{a fraction} \). Integrate both sides and then put \( (x+y) \) back for 't'.

🎯 Exam Tip: For rational functions in 't' (like \( \frac{2t+3}{3t+4} \)), adjust the numerator by expressing it in terms of the denominator. This technique, \( \frac{At+B}{Ct+D} = \frac{A}{C} + \frac{BC-AD}{C(Ct+D)} \), allows for easier integration into linear and logarithmic terms.