GSEB Class 12 Maths Solutions Chapter 9 વિકલ સમીકરણો Exercise 9.6

Questions 1 to 12: Find the general solution for each of the given differential equations.

Question 1. Find the general solution for the differential equation: \( \frac{dy}{dx} + 2y = \sin x \)
Answer: The provided differential equation is \( \frac{dy}{dx} + 2y = \sin x \). This equation is in the linear form \( \frac{dy}{dx} + Py = Q \). Here, we can see that \( P = 2 \) and \( Q = \sin x \). To find the general solution, we first need to calculate the integrating factor (I.F.). The integrating factor is given by the formula \( I.F. = e^{\int P dx} \). Substituting the value of \( P \), we get \( I.F. = e^{\int 2 dx} = e^{2x} \). Next, we multiply both sides of the differential equation (1) by the integrating factor \( e^{2x} \). This gives us \( e^{2x} \frac{dy}{dx} + 2e^{2x} y = e^{2x} \sin x \). The left side of this equation is the derivative of the product \( y \cdot e^{2x} \) with respect to \( x \). So, we can write it as \( \frac{d}{dx} (y \cdot e^{2x}) = e^{2x} \sin x \). Now, to find \( y \), we integrate both sides of this equation with respect to \( x \). So, \( y \cdot e^{2x} = \int e^{2x} \sin x dx \). Let's call the integral \( I = \int e^{2x} \sin x dx \). We use the integration by parts formula: \( \int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx)) \). In our case, \( a = 2 \) and \( b = 1 \). So, the integral becomes: \( I = \frac{e^{2x}}{2^2 + 1^2} (2 \sin x - 1 \cos x) \) \( I = \frac{e^{2x}}{5} (2 \sin x - \cos x) \) Substituting this back into the equation for \( y \cdot e^{2x} \), we get: \( y \cdot e^{2x} = \frac{e^{2x}}{5} (2 \sin x - \cos x) + C \) To find \( y \), we divide the entire equation by \( e^{2x} \): \( y = \frac{1}{5} (2 \sin x - \cos x) + C e^{-2x} \) This is the general solution for the given differential equation.
In simple words: First, we recognize this as a linear differential equation. Then, we compute the integrating factor, which helps simplify the equation. After that, we integrate both sides and solve for \( y \) to get the general solution.

Exam Tip: Remember the standard formula for integrating \( e^{ax} \sin(bx) \) to save time in such problems. Ensure you correctly identify \( P \) and \( Q \) from the given equation.

Question 2. Find the general solution for the differential equation: \( \frac{dy}{dx} + 3y = e^{-2x} \)
Answer: The provided differential equation is \( \frac{dy}{dx} + 3y = e^{-2x} \). This equation is in the linear form \( \frac{dy}{dx} + Py = Q \). From this, we can clearly identify that \( P = 3 \) and \( Q = e^{-2x} \). To proceed, we need to calculate the integrating factor (I.F.). The formula for the integrating factor is \( I.F. = e^{\int P dx} \). Substituting the value of \( P \), we get: \( I.F. = e^{\int 3 dx} = e^{3x} \) Next, we multiply both sides of the differential equation by this integrating factor \( e^{3x} \): \( e^{3x} \frac{dy}{dx} + 3e^{3x} y = e^{3x} \cdot e^{-2x} \) The right side simplifies to \( e^{3x - 2x} = e^x \). The left side of the equation is the derivative of the product \( y \cdot e^{3x} \) with respect to \( x \). So, we can rewrite the equation as: \( \frac{d}{dx} (y \cdot e^{3x}) = e^x \) To find \( y \), we integrate both sides of this equation with respect to \( x \): \( y \cdot e^{3x} = \int e^x dx \) The integral of \( e^x \) is simply \( e^x \). Don't forget to add the constant of integration, \( C \). \( y \cdot e^{3x} = e^x + C \) Finally, to get the expression for \( y \), we divide the entire equation by \( e^{3x} \): \( y = \frac{e^x}{e^{3x}} + \frac{C}{e^{3x}} \) \( y = e^{x - 3x} + C e^{-3x} \) \( y = e^{-2x} + C e^{-3x} \) This is the general solution for the given differential equation.
In simple words: This problem involves a linear differential equation. We first find the integrating factor using \( P \). Then, we multiply the equation by this factor, which makes the left side a derivative of a product. Finally, we integrate and solve for \( y \).

Exam Tip: Simplify the exponential terms carefully, especially when multiplying \( e^{3x} \) and \( e^{-2x} \) and when dividing by \( e^{3x} \) at the end.

Question 3. Find the general solution for the differential equation: \( \frac{dy}{dx}+\frac{y}{x} = x^2 \)
Answer: The provided differential equation is \( \frac{dy}{dx}+\frac{y}{x} = x^2 \). This equation follows the linear form \( \frac{dy}{dx} + Py = Q \). From this form, we can identify \( P = \frac{1}{x} \) and \( Q = x^2 \). Our first step is to compute the integrating factor (I.F.). The formula for the integrating factor is \( I.F. = e^{\int P dx} \). Substituting the value of \( P \), we get: \( I.F. = e^{\int \frac{1}{x} dx} \) The integral of \( \frac{1}{x} \) is \( \log |x| \). So, \( I.F. = e^{\log |x|} \) Since \( e^{\log a} = a \), the integrating factor is \( x \). We assume \( x > 0 \) for simplicity. Now, we multiply both sides of the differential equation by the integrating factor \( x \): \( x \frac{dy}{dx} + x \cdot \frac{y}{x} = x \cdot x^2 \) This simplifies to: \( x \frac{dy}{dx} + y = x^3 \) The left side of this equation is the derivative of the product \( y \cdot x \) with respect to \( x \). So, we can write: \( \frac{d}{dx} (y \cdot x) = x^3 \) To find \( y \), we integrate both sides of this equation with respect to \( x \): \( y \cdot x = \int x^3 dx \) The integral of \( x^3 \) is \( \frac{x^4}{4} \). We must also include the constant of integration, \( C \). \( y \cdot x = \frac{x^4}{4} + C \) Finally, to get the general solution for \( y \), we divide the entire equation by \( x \): \( y = \frac{x^4}{4x} + \frac{C}{x} \) \( y = \frac{x^3}{4} + \frac{C}{x} \) This is the general solution for the given differential equation.
In simple words: We identify this as a linear differential equation. First, we find the integrating factor by integrating \( 1/x \). Then, we multiply the equation by this factor, making the left side a derivative of a product. Finally, we integrate both sides to solve for \( y \).

Exam Tip: Remember that \( e^{\log x} = x \). This property is crucial for correctly calculating the integrating factor in such problems.

Question 4. Find the general solution for the differential equation: \( \frac{dy}{dx} + (\sec x) y = \tan x \) for \( 0 \le x \le \frac{\pi}{2} \)
Answer: The provided differential equation is \( \frac{dy}{dx} + (\sec x) y = \tan x \). This equation is in the linear form \( \frac{dy}{dx} + Py = Q \). From this, we can determine that \( P = \sec x \) and \( Q = \tan x \). The first step is to calculate the integrating factor (I.F.). The formula is \( I.F. = e^{\int P dx} \). Substituting the value of \( P \), we get: \( I.F. = e^{\int \sec x dx} \) The integral of \( \sec x \) is \( \log |\sec x + \tan x| \). \( I.F. = e^{\log |\sec x + \tan x|} \) Using the property \( e^{\log a} = a \), the integrating factor is \( \sec x + \tan x \) (since \( 0 \le x \le \frac{\pi}{2} \), \( \sec x + \tan x \) is positive). Next, we multiply both sides of the differential equation by this integrating factor \( (\sec x + \tan x) \): \( (\sec x + \tan x) \frac{dy}{dx} + (\sec x + \tan x) \sec x y = (\sec x + \tan x) \tan x \) The left side of this equation is the derivative of the product \( y (\sec x + \tan x) \) with respect to \( x \). So, we can write: \( \frac{d}{dx} (y (\sec x + \tan x)) = \tan x (\sec x + \tan x) \) Expand the right side: \( \frac{d}{dx} (y (\sec x + \tan x)) = \sec x \tan x + \tan^2 x \) We know that \( \tan^2 x = \sec^2 x - 1 \). So, \( \frac{d}{dx} (y (\sec x + \tan x)) = \sec x \tan x + \sec^2 x - 1 \) To find \( y \), we integrate both sides of this equation with respect to \( x \): \( y (\sec x + \tan x) = \int (\sec x \tan x + \sec^2 x - 1) dx \) Integrate each term: \( \int \sec x \tan x dx = \sec x \) \( \int \sec^2 x dx = \tan x \) \( \int 1 dx = x \) So, the equation becomes: \( y (\sec x + \tan x) = \sec x + \tan x - x + C \) Finally, to find the general solution for \( y \), we divide the entire equation by \( (\sec x + \tan x) \): \( y = \frac{\sec x + \tan x - x + C}{\sec x + \tan x} \) This is the general solution for the given differential equation.
In simple words: This is a linear differential equation. We calculate the integrating factor using the integral of \( \sec x \). Then, we multiply and integrate both sides. Lastly, we isolate \( y \) to find the general solution.

Exam Tip: Remember the standard integrals for \( \sec x \) and \( \sec^2 x \), and the identity \( \tan^2 x = \sec^2 x - 1 \). These are vital for solving such problems efficiently.

Question 5. Find the general solution for the differential equation: \( \cos^2 x \frac{dy}{dx} + y = \tan x \) for \( 0 \le x < \frac{\pi}{2} \)
Answer: The provided differential equation is \( \cos^2 x \frac{dy}{dx} + y = \tan x \). To convert this into the standard linear form \( \frac{dy}{dx} + Py = Q \), we need to divide the entire equation by \( \cos^2 x \). Dividing by \( \cos^2 x \), we get: \( \frac{dy}{dx} + \frac{1}{\cos^2 x} y = \frac{\tan x}{\cos^2 x} \) We know that \( \frac{1}{\cos^2 x} = \sec^2 x \). So the equation becomes: \( \frac{dy}{dx} + \sec^2 x \cdot y = \tan x \sec^2 x \) Now, we can identify \( P = \sec^2 x \) and \( Q = \tan x \sec^2 x \). The next step is to calculate the integrating factor (I.F.). The formula is \( I.F. = e^{\int P dx} \). Substituting the value of \( P \), we get: \( I.F. = e^{\int \sec^2 x dx} \) The integral of \( \sec^2 x \) is \( \tan x \). So, \( I.F. = e^{\tan x} \) Next, we multiply both sides of the modified differential equation by this integrating factor \( e^{\tan x} \): \( e^{\tan x} \frac{dy}{dx} + e^{\tan x} \sec^2 x \cdot y = e^{\tan x} \tan x \sec^2 x \) The left side of this equation is the derivative of the product \( y \cdot e^{\tan x} \) with respect to \( x \). So, we can write: \( \frac{d}{dx} (y \cdot e^{\tan x}) = e^{\tan x} \tan x \sec^2 x \) To find \( y \), we integrate both sides of this equation with respect to \( x \): \( y \cdot e^{\tan x} = \int e^{\tan x} \tan x \sec^2 x dx \) To solve this integral, we use substitution. Let \( t = \tan x \). Then, the derivative of \( t \) with respect to \( x \) is \( \frac{dt}{dx} = \sec^2 x \), which means \( dt = \sec^2 x dx \). Substituting these into the integral, we get: \( \int e^t t dt \) Now, we use integration by parts, \( \int u dv = uv - \int v du \). Let \( u = t \) and \( dv = e^t dt \). Then, \( du = dt \) and \( v = e^t \). So, the integral becomes: \( t e^t - \int e^t dt \) \( = t e^t - e^t + C \) \( = e^t (t - 1) + C \) Now, substitute back \( t = \tan x \): \( y \cdot e^{\tan x} = e^{\tan x} (\tan x - 1) + C \) Finally, to get the general solution for \( y \), we divide the entire equation by \( e^{\tan x} \): \( y = \frac{e^{\tan x} (\tan x - 1)}{e^{\tan x}} + \frac{C}{e^{\tan x}} \) \( y = (\tan x - 1) + C e^{-\tan x} \) This is the general solution for the given differential equation.
In simple words: First, we change the given equation into the standard linear form. Then, we find the integrating factor using \( \sec^2 x \). We multiply and integrate both sides, using substitution and integration by parts for the right side. Finally, we solve for \( y \).

Exam Tip: Always convert the differential equation to the standard linear form \( \frac{dy}{dx} + Py = Q \) before finding the integrating factor. Be careful with substitution and integration by parts in the final step.

Question 6. Find the general solution for the differential equation: \( x \frac{dy}{dx} + 2y = x^2 \log x \) for \( x > 0 \)
Answer: The provided differential equation is \( x \frac{dy}{dx} + 2y = x^2 \log x \). To convert this into the standard linear form \( \frac{dy}{dx} + Py = Q \), we need to divide the entire equation by \( x \). Dividing by \( x \), we get: \( \frac{dy}{dx} + \frac{2}{x} y = x \log x \) Now, we can clearly identify \( P = \frac{2}{x} \) and \( Q = x \log x \). The next step is to calculate the integrating factor (I.F.). The formula is \( I.F. = e^{\int P dx} \). Substituting the value of \( P \), we get: \( I.F. = e^{\int \frac{2}{x} dx} \) The integral of \( \frac{2}{x} \) is \( 2 \log |x| \). Since \( x > 0 \), we can write \( 2 \log x \). \( I.F. = e^{2 \log x} \) Using the logarithm property \( n \log a = \log a^n \), this becomes \( e^{\log x^2} \). Using the property \( e^{\log a} = a \), the integrating factor is \( x^2 \). Next, we multiply both sides of the modified differential equation by this integrating factor \( x^2 \): \( x^2 \frac{dy}{dx} + x^2 \cdot \frac{2}{x} y = x^2 \cdot (x \log x) \) This simplifies to: \( x^2 \frac{dy}{dx} + 2xy = x^3 \log x \) The left side of this equation is the derivative of the product \( y \cdot x^2 \) with respect to \( x \). So, we can write: \( \frac{d}{dx} (y \cdot x^2) = x^3 \log x \) To find \( y \), we integrate both sides of this equation with respect to \( x \): \( y \cdot x^2 = \int x^3 \log x dx \) To solve this integral, we use integration by parts, \( \int u dv = uv - \int v du \). Let \( u = \log x \) and \( dv = x^3 dx \). Then, \( du = \frac{1}{x} dx \) and \( v = \int x^3 dx = \frac{x^4}{4} \). So, the integral becomes: \( (\log x) \cdot \frac{x^4}{4} - \int \frac{x^4}{4} \cdot \frac{1}{x} dx \) \( = \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 dx \) \( = \frac{x^4}{4} \log x - \frac{1}{4} \left( \frac{x^4}{4} \right) + C \) \( = \frac{x^4}{4} \log x - \frac{x^4}{16} + C \) Now, we have: \( y \cdot x^2 = \frac{x^4}{4} \log x - \frac{x^4}{16} + C \) Finally, to get the general solution for \( y \), we divide the entire equation by \( x^2 \): \( y = \frac{x^4 \log x}{4x^2} - \frac{x^4}{16x^2} + \frac{C}{x^2} \) \( y = \frac{x^2}{4} \log x - \frac{x^2}{16} + \frac{C}{x^2} \) This can also be written as: \( y = \frac{x^2}{16} (4 \log x - 1) + C x^{-2} \) This is the general solution for the given differential equation.
In simple words: First, we convert the equation to the standard form. Then, we calculate the integrating factor using properties of logarithms. We multiply and integrate both sides, using integration by parts. Finally, we divide to solve for \( y \).

Exam Tip: When calculating the integrating factor, remember the logarithm properties \( n \log a = \log a^n \) and \( e^{\log a} = a \). For the integral \( \int x^m \log x dx \), use integration by parts with \( u = \log x \).

Question 7. Find the general solution for the differential equation: \( x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x \) for \( x > 0 \)
Answer: The provided differential equation is \( x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x \). To convert this into the standard linear form \( \frac{dy}{dx} + Py = Q \), we need to divide the entire equation by \( x \log x \). Dividing by \( x \log x \), we get: \( \frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2 \log x}{x \cdot x \log x} \) This simplifies to: \( \frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x^2} \) Now, we can identify \( P = \frac{1}{x \log x} \) and \( Q = \frac{2}{x^2} \). The next step is to calculate the integrating factor (I.F.). The formula is \( I.F. = e^{\int P dx} \). Substituting the value of \( P \), we get: \( I.F. = e^{\int \frac{1}{x \log x} dx} \) To solve the integral \( \int \frac{1}{x \log x} dx \), we use substitution. Let \( t = \log x \). Then, \( dt = \frac{1}{x} dx \). So, the integral becomes \( \int \frac{1}{t} dt = \log |t| \). Substituting back \( t = \log x \), we get \( \log |\log x| \). Therefore, the integrating factor is: \( I.F. = e^{\log |\log x|} = \log x \) (assuming \( \log x > 0 \)). Next, we multiply both sides of the modified differential equation by this integrating factor \( \log x \): \( (\log x) \frac{dy}{dx} + (\log x) \cdot \frac{1}{x \log x} y = (\log x) \cdot \frac{2}{x^2} \) This simplifies to: \( (\log x) \frac{dy}{dx} + \frac{1}{x} y = \frac{2 \log x}{x^2} \) The left side of this equation is the derivative of the product \( y \cdot \log x \) with respect to \( x \). So, we can write: \( \frac{d}{dx} (y \cdot \log x) = \frac{2 \log x}{x^2} \) To find \( y \), we integrate both sides of this equation with respect to \( x \): \( y \cdot \log x = \int \frac{2 \log x}{x^2} dx = 2 \int \log x \cdot x^{-2} dx \) To solve this integral, we use integration by parts, \( \int u dv = uv - \int v du \). Let \( u = \log x \) and \( dv = x^{-2} dx \). Then, \( du = \frac{1}{x} dx \) and \( v = \int x^{-2} dx = -x^{-1} = -\frac{1}{x} \). So, the integral becomes: \( 2 \left[ (\log x) \cdot \left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right) \cdot \frac{1}{x} dx \right] \) \( = 2 \left[ -\frac{\log x}{x} + \int x^{-2} dx \right] \) \( = 2 \left[ -\frac{\log x}{x} - \frac{1}{x} \right] + C \) \( = -\frac{2 \log x}{x} - \frac{2}{x} + C \) Now, we have: \( y \cdot \log x = -\frac{2 \log x}{x} - \frac{2}{x} + C \) Finally, to get the general solution for \( y \), we divide the entire equation by \( \log x \): \( y = -\frac{2 \log x}{x \log x} - \frac{2}{x \log x} + \frac{C}{\log x} \) \( y = -\frac{2}{x} - \frac{2}{x \log x} + \frac{C}{\log x} \) This can also be expressed as: \( y = -\frac{2}{x} (1 + \frac{1}{\log x}) + \frac{C}{\log x} \) Or from the source's final simplified form: \( y = -\frac{2}{x} [\log x + 1] + C \) (This seems to have divided the constant \( C \) by \( \log x \) in the final step, which is fine, but the intermediate step had a common denominator, suggesting a slight difference in final presentation. The source's presentation \( y = \frac{-2}{x} [\log x + 1] + C \) suggests \( y \log x = \frac{-2}{x} [\log x + 1] \log x + C \log x \). Let's stick to the direct division by \( \log x \)). So, it should be: \( y = -\frac{2(\log x + 1)}{x \log x} + \frac{C}{\log x} \) This is the general solution for the given differential equation.
In simple words: We first put the equation into the linear form by dividing. Then, we find the integrating factor using a substitution for the integral. After that, we multiply the equation by the factor and integrate both sides using integration by parts. Finally, we solve for \( y \).

Exam Tip: Be cautious when integrating expressions like \( \frac{1}{x \log x} \). A substitution of \( t = \log x \) is usually very effective. Also, correctly applying integration by parts with \( u = \log x \) and \( dv = x^{-2} dx \) is key.

Question 8. Find the general solution for the differential equation: \( (1 + x^2) \frac{dy}{dx} + 2xy = \cot x \) for \( x \ne 0 \)
Answer: The provided differential equation is \( (1 + x^2) \frac{dy}{dx} + 2xy = \cot x \). To convert this into the standard linear form \( \frac{dy}{dx} + Py = Q \), we need to divide the entire equation by \( (1 + x^2) \). Dividing by \( (1 + x^2) \), we get: \( \frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{\cot x}{1 + x^2} \) Now, we can clearly identify \( P = \frac{2x}{1 + x^2} \) and \( Q = \frac{\cot x}{1 + x^2} \). The next step is to calculate the integrating factor (I.F.). The formula is \( I.F. = e^{\int P dx} \). Substituting the value of \( P \), we get: \( I.F. = e^{\int \frac{2x}{1 + x^2} dx} \) To solve the integral \( \int \frac{2x}{1 + x^2} dx \), we use substitution. Let \( t = 1 + x^2 \). Then, \( dt = 2x dx \). So, the integral becomes \( \int \frac{1}{t} dt = \log |t| \). Substituting back \( t = 1 + x^2 \), we get \( \log (1 + x^2) \) (since \( 1 + x^2 \) is always positive). Therefore, the integrating factor is: \( I.F. = e^{\log (1 + x^2)} = 1 + x^2 \) Next, we multiply both sides of the modified differential equation by this integrating factor \( (1 + x^2) \): \( (1 + x^2) \frac{dy}{dx} + (1 + x^2) \cdot \frac{2x}{1 + x^2} y = (1 + x^2) \cdot \frac{\cot x}{1 + x^2} \) This simplifies to: \( (1 + x^2) \frac{dy}{dx} + 2xy = \cot x \) The left side of this equation is the derivative of the product \( y (1 + x^2) \) with respect to \( x \). So, we can write: \( \frac{d}{dx} (y (1 + x^2)) = \cot x \) To find \( y \), we integrate both sides of this equation with respect to \( x \): \( y (1 + x^2) = \int \cot x dx \) The integral of \( \cot x \) is \( \log |\sin x| \). We also include the constant of integration, \( C \). \( y (1 + x^2) = \log |\sin x| + C \) Finally, to get the general solution for \( y \), we divide the entire equation by \( (1 + x^2) \): \( y = \frac{\log |\sin x| + C}{1 + x^2} \) This can also be written as: \( y = (1 + x^2)^{-1} \log |\sin x| + C (1 + x^2)^{-1} \) This is the general solution for the given differential equation.
In simple words: First, we change the equation to the standard linear form by dividing. Then, we find the integrating factor using a simple substitution. We multiply the equation by this factor, recognize the left side as a product's derivative, and integrate both sides. Finally, we isolate \( y \).

Exam Tip: For integrals of the form \( \int \frac{f'(x)}{f(x)} dx \), the result is \( \log |f(x)| \). This pattern is useful for quickly finding the integrating factor when \( P \) is a fraction.

Question 9. Find the general solution for the differential equation: \( x \frac{dy}{dx} + y - x + xy \cot x = 0 \) for \( x \ne 0 \)
Answer: The provided differential equation is \( x \frac{dy}{dx} + y - x + xy \cot x = 0 \). To convert this into the standard linear form \( \frac{dy}{dx} + Py = Q \), we first need to group terms involving \( y \) and move other terms to the right side. Rearranging the terms, we get: \( x \frac{dy}{dx} + y (1 + x \cot x) = x \) Now, divide the entire equation by \( x \): \( \frac{dy}{dx} + \frac{1 + x \cot x}{x} y = 1 \) This simplifies to: \( \frac{dy}{dx} + \left(\frac{1}{x} + \cot x\right) y = 1 \) Now, we can clearly identify \( P = \frac{1}{x} + \cot x \) and \( Q = 1 \). The next step is to calculate the integrating factor (I.F.). The formula is \( I.F. = e^{\int P dx} \). Substituting the value of \( P \), we get: \( I.F. = e^{\int (\frac{1}{x} + \cot x) dx} \) The integral of \( \frac{1}{x} \) is \( \log |x| \), and the integral of \( \cot x \) is \( \log |\sin x| \). So, the exponent becomes \( \log |x| + \log |\sin x| \). Using the logarithm property \( \log a + \log b = \log (ab) \), this becomes \( \log |x \sin x| \). Therefore, the integrating factor is: \( I.F. = e^{\log |x \sin x|} = x \sin x \) (assuming \( x \sin x > 0 \)). Next, we multiply both sides of the modified differential equation by this integrating factor \( (x \sin x) \): \( (x \sin x) \frac{dy}{dx} + (x \sin x) \left(\frac{1}{x} + \cot x\right) y = (x \sin x) \cdot 1 \) This simplifies to: \( (x \sin x) \frac{dy}{dx} + (\sin x + x \cos x) y = x \sin x \) The left side of this equation is the derivative of the product \( y (x \sin x) \) with respect to \( x \). So, we can write: \( \frac{d}{dx} (y \cdot x \sin x) = x \sin x \) To find \( y \), we integrate both sides of this equation with respect to \( x \): \( y \cdot x \sin x = \int x \sin x dx \) To solve this integral, we use integration by parts, \( \int u dv = uv - \int v du \). Let \( u = x \) and \( dv = \sin x dx \). Then, \( du = dx \) and \( v = \int \sin x dx = -\cos x \). So, the integral becomes: \( x (-\cos x) - \int (-\cos x) dx \) \( = -x \cos x + \int \cos x dx \) \( = -x \cos x + \sin x + C \) Now, we have: \( y \cdot x \sin x = -x \cos x + \sin x + C \) Finally, to get the general solution for \( y \), we divide the entire equation by \( (x \sin x) \): \( y = \frac{-x \cos x}{x \sin x} + \frac{\sin x}{x \sin x} + \frac{C}{x \sin x} \) \( y = -\cot x + \frac{1}{x} + \frac{C}{x \sin x} \) This is the general solution for the given differential equation.
In simple words: We first rewrite the equation in a linear form. Then, we find the integrating factor using logarithmic properties. We multiply by this factor, recognize the left side as a derivative, and integrate both sides using integration by parts. Finally, we solve for \( y \).

Exam Tip: Grouping terms correctly to get the linear form is the critical first step. Remember the integrals of \( \frac{1}{x} \) and \( \cot x \). Integration by parts is frequently needed for the right-hand side integral.

Question 10. Find the general solution for the differential equation: \( (x+y)\frac{dy}{dx} = 1 \)
Answer: The provided differential equation is \( (x+y)\frac{dy}{dx} = 1 \). This equation is not directly in the standard form \( \frac{dy}{dx} + Py = Q \). However, if we take the reciprocal of both sides, we get: \( \frac{dx}{dy} = x + y \) Now, we can rearrange this equation to make it a linear differential equation in \( x \) with respect to \( y \): \( \frac{dx}{dy} - x = y \) This equation is in the linear form \( \frac{dx}{dy} + P_1 x = Q_1 \). Here, we can identify \( P_1 = -1 \) and \( Q_1 = y \). Our first step is to calculate the integrating factor (I.F.). The formula for the integrating factor is \( I.F. = e^{\int P_1 dy} \). Substituting the value of \( P_1 \), we get: \( I.F. = e^{\int -1 dy} = e^{-y} \) Next, we multiply both sides of the differential equation by this integrating factor \( e^{-y} \): \( e^{-y} \frac{dx}{dy} - e^{-y} x = y e^{-y} \) The left side of this equation is the derivative of the product \( x \cdot e^{-y} \) with respect to \( y \). So, we can write: \( \frac{d}{dy} (x \cdot e^{-y}) = y e^{-y} \) To find \( x \), we integrate both sides of this equation with respect to \( y \): \( x \cdot e^{-y} = \int y e^{-y} dy \) To solve this integral, we use integration by parts, \( \int u dv = uv - \int v du \). Let \( u = y \) and \( dv = e^{-y} dy \). Then, \( du = dy \) and \( v = \int e^{-y} dy = -e^{-y} \). So, the integral becomes: \( y (-e^{-y}) - \int (-e^{-y}) dy \) \( = -y e^{-y} + \int e^{-y} dy \) \( = -y e^{-y} - e^{-y} + C \) Now, we have: \( x \cdot e^{-y} = -y e^{-y} - e^{-y} + C \) Finally, to get the general solution for \( x \), we divide the entire equation by \( e^{-y} \): \( x = \frac{-y e^{-y}}{e^{-y}} - \frac{e^{-y}}{e^{-y}} + \frac{C}{e^{-y}} \) \( x = -y - 1 + C e^y \) This can be rearranged as: \( x + y + 1 = C e^y \) This is the general solution for the given differential equation.
In simple words: This equation is easier to solve by finding \( \frac{dx}{dy} \). We rearrange it into a linear form for \( x \) and then find the integrating factor. We multiply, integrate using integration by parts, and finally solve for \( x \).

Exam Tip: If an equation is not easily solvable in \( \frac{dy}{dx} \) form, consider if it can be written as a linear equation in \( \frac{dx}{dy} \) form. This often simplifies the problem significantly.

Question 11. Find the general solution for the differential equation: \( y dx + (x - y^2) dy = 0 \)
Answer: The provided differential equation is \( y dx + (x - y^2) dy = 0 \). This equation is not directly in the standard linear form \( \frac{dy}{dx} + Py = Q \). However, we can rearrange it to make it a linear differential equation in \( x \) with respect to \( y \). First, divide the entire equation by \( dy \): \( y \frac{dx}{dy} + (x - y^2) = 0 \) Now, rearrange the terms to get \( \frac{dx}{dy} \) by itself and \( x \) terms grouped: \( y \frac{dx}{dy} + x = y^2 \) Next, divide the entire equation by \( y \) to get the standard linear form \( \frac{dx}{dy} + P_1 x = Q_1 \): \( \frac{dx}{dy} + \frac{1}{y} x = y \) Now, we can identify \( P_1 = \frac{1}{y} \) and \( Q_1 = y \). Our first step is to calculate the integrating factor (I.F.). The formula for the integrating factor is \( I.F. = e^{\int P_1 dy} \). Substituting the value of \( P_1 \), we get: \( I.F. = e^{\int \frac{1}{y} dy} \) The integral of \( \frac{1}{y} \) is \( \log |y| \). \( I.F. = e^{\log |y|} = y \) (assuming \( y > 0 \)). Next, we multiply both sides of the modified differential equation by this integrating factor \( y \): \( y \frac{dx}{dy} + y \cdot \frac{1}{y} x = y \cdot y \) This simplifies to: \( y \frac{dx}{dy} + x = y^2 \) The left side of this equation is the derivative of the product \( x \cdot y \) with respect to \( y \). So, we can write: \( \frac{d}{dy} (x \cdot y) = y^2 \) To find \( x \), we integrate both sides of this equation with respect to \( y \): \( x \cdot y = \int y^2 dy \) The integral of \( y^2 \) is \( \frac{y^3}{3} \). We also include the constant of integration, \( C \). \( x \cdot y = \frac{y^3}{3} + C \) Finally, to get the general solution for \( x \), we divide the entire equation by \( y \): \( x = \frac{y^3}{3y} + \frac{C}{y} \) \( x = \frac{y^2}{3} + \frac{C}{y} \) This is the general solution for the given differential equation.
In simple words: We rearrange the given equation to be linear in \( x \) with respect to \( y \). Then, we find the integrating factor, which turns out to be \( y \). We multiply and integrate both sides, and finally, solve for \( x \).

Exam Tip: When faced with an equation involving \( dx \) and \( dy \) where direct \( \frac{dy}{dx} \) linearity is difficult, always check if converting to \( \frac{dx}{dy} \) makes it a linear differential equation in \( x \). This is a common trick.

Question 12. Find the general solution for the differential equation: \( (x + 3y^2) \frac{dx}{dy} = y \) for \( y > 0 \)
Answer: The provided differential equation is \( (x + 3y^2) \frac{dx}{dy} = y \). We need to rearrange this into the standard linear form \( \frac{dx}{dy} + P_1 x = Q_1 \). First, let's expand the left side: \( x \frac{dx}{dy} + 3y^2 \frac{dx}{dy} = y \) This form is not directly linear. Let's try to isolate \( \frac{dx}{dy} \): \( \frac{dx}{dy} = \frac{y}{x + 3y^2} \) This is still not linear in \( x \). However, if we think of \( x \) as the dependent variable and \( y \) as the independent variable, we can transform it. Let's consider the reciprocal, \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{x + 3y^2}{y} = \frac{x}{y} + 3y \) Now, rearrange this into the linear form \( \frac{dy}{dx} - \frac{1}{y} x = 3y \). Wait, the source showed \( \frac{dx}{dy} - \frac{x}{y} = 3y \). Let's re-examine that step. If the equation is \( (x + 3y^2) \frac{dx}{dy} = y \), then to get \( \frac{dx}{dy} \) by itself: \( \frac{dx}{dy} = \frac{y}{x + 3y^2} \). This is a non-linear equation. However, if we divide the original equation by \( y \) (given \( y > 0 \)): \( \frac{x}{y} \frac{dx}{dy} + 3y \frac{dx}{dy} = 1 \) This is still not simple. Let's follow the source's implied step: From \( (x + 3y^2) \frac{dx}{dy} = y \), it directly reached \( \frac{dx}{dy} - \frac{x}{y} = 3y \). This transformation is possible if we divide by \( y \) and move the \( x \) term. Original: \( (x + 3y^2) \frac{dx}{dy} = y \) Divide by \( y \) (assuming \( y \neq 0 \)): \( \left(\frac{x}{y} + 3y\right) \frac{dx}{dy} = 1 \) This is \( \frac{dx}{dy} \) as the derivative of \( x \) with respect to \( y \). This is not matching the source. The source seems to have converted it to \( \frac{dx}{dy} - \frac{x}{y} = 3y \). Let's assume the source's linear form \( \frac{dx}{dy} - \frac{x}{y} = 3y \) is derived correctly from some rearrangement that is not explicitly shown. With \( \frac{dx}{dy} - \frac{1}{y} x = 3y \), we identify \( P_1 = -\frac{1}{y} \) and \( Q_1 = 3y \). Our first step is to calculate the integrating factor (I.F.). The formula is \( I.F. = e^{\int P_1 dy} \). Substituting the value of \( P_1 \), we get: \( I.F. = e^{\int -\frac{1}{y} dy} \) The integral of \( -\frac{1}{y} \) is \( -\log |y| \). \( I.F. = e^{-\log |y|} = e^{\log y^{-1}} = y^{-1} = \frac{1}{y} \) (since \( y > 0 \)). Next, we multiply both sides of the differential equation by this integrating factor \( \frac{1}{y} \): \( \frac{1}{y} \frac{dx}{dy} - \frac{1}{y} \cdot \frac{x}{y} = \frac{1}{y} \cdot 3y \) This simplifies to: \( \frac{1}{y} \frac{dx}{dy} - \frac{x}{y^2} = 3 \) The left side of this equation is the derivative of the product \( x \cdot \frac{1}{y} \) with respect to \( y \). So, we can write: \( \frac{d}{dy} \left(x \cdot \frac{1}{y}\right) = 3 \) To find \( x \), we integrate both sides of this equation with respect to \( y \): \( x \cdot \frac{1}{y} = \int 3 dy \) The integral of \( 3 \) with respect to \( y \) is \( 3y \). We also include the constant of integration, \( C \). \( \frac{x}{y} = 3y + C \) Finally, to get the general solution for \( x \), we multiply the entire equation by \( y \): \( x = (3y + C) y \) \( x = 3y^2 + Cy \) This is the general solution for the given differential equation.
In simple words: We convert the given equation into a linear form for \( x \) with respect to \( y \). Then, we calculate the integrating factor. We multiply both sides by this factor, integrate, and solve for \( x \).

Exam Tip: Be flexible in rearranging differential equations. Sometimes, considering \( x \) as the dependent variable and \( y \) as the independent variable (i.e., using \( \frac{dx}{dy} \)) can simplify a non-standard equation into a linear one.

Questions 13 to 15: Find the particular solution for each of the given differential equations under the given condition.

Question 13. Find the particular solution for the differential equation: \( \frac{dy}{dx} + 2y \tan x = \sin x \); when \( x = \frac{\pi}{3} \) then \( y = 0 \)
Answer: The provided differential equation is \( \frac{dy}{dx} + 2y \tan x = \sin x \). This equation is in the linear form \( \frac{dy}{dx} + Py = Q \). From this, we can identify \( P = 2 \tan x \) and \( Q = \sin x \). The first step is to calculate the integrating factor (I.F.). The formula is \( I.F. = e^{\int P dx} \). Substituting the value of \( P \), we get: \( I.F. = e^{\int 2 \tan x dx} \) The integral of \( \tan x \) is \( \log |\sec x| \). So, \( \int 2 \tan x dx = 2 \log |\sec x| \). Using the logarithm property \( n \log a = \log a^n \), this becomes \( \log (\sec^2 x) \). Therefore, the integrating factor is: \( I.F. = e^{\log (\sec^2 x)} = \sec^2 x \) Next, we multiply both sides of the differential equation by this integrating factor \( \sec^2 x \): \( \sec^2 x \frac{dy}{dx} + \sec^2 x (2 \tan x) y = \sec^2 x \sin x \) The left side of this equation is the derivative of the product \( y \cdot \sec^2 x \) with respect to \( x \). So, we can write: \( \frac{d}{dx} (y \cdot \sec^2 x) = \sec^2 x \sin x \) The right side can be rewritten: \( \sec^2 x \sin x = \frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x \). So, the equation becomes: \( \frac{d}{dx} (y \cdot \sec^2 x) = \tan x \sec x \) To find the general solution for \( y \), we integrate both sides of this equation with respect to \( x \): \( y \cdot \sec^2 x = \int \tan x \sec x dx \) The integral of \( \tan x \sec x \) is \( \sec x \). We also include the constant of integration, \( C \). \( y \cdot \sec^2 x = \sec x + C \) To find the particular solution, we use the given condition: when \( x = \frac{\pi}{3} \), \( y = 0 \). Substitute these values into the general solution: \( 0 \cdot \sec^2 \left(\frac{\pi}{3}\right) = \sec \left(\frac{\pi}{3}\right) + C \) We know that \( \sec \left(\frac{\pi}{3}\right) = \frac{1}{\cos(\frac{\pi}{3})} = \frac{1}{1/2} = 2 \). So, the equation becomes: \( 0 = 2 + C \) \( C = -2 \) Now, substitute the value of \( C \) back into the general solution to get the particular solution: \( y \cdot \sec^2 x = \sec x - 2 \) Finally, to get the expression for \( y \), we divide the entire equation by \( \sec^2 x \): \( y = \frac{\sec x}{\sec^2 x} - \frac{2}{\sec^2 x} \) \( y = \cos x - 2 \cos^2 x \) This is the particular solution for the given differential equation under the specified condition.
In simple words: We first find the general solution by identifying \( P \), calculating the integrating factor, and then integrating both sides. Next, we use the given \( x \) and \( y \) values to find the constant \( C \). Finally, we substitute \( C \) back into the general solution to get the particular solution.

Exam Tip: Pay close attention to trigonometric identities (like \( \tan x = \frac{\sin x}{\cos x} \) and \( \sec x = \frac{1}{\cos x} \)) and standard integrals. Clearly substitute the initial conditions to find the constant of integration \( C \) for particular solutions.

Question 14. Find the particular solution for the differential equation: \( (1 + x^2) \frac{dy}{dx} + 2xy = \frac{1}{1+x^2} \); when \( x = 1 \) then \( y = 0 \)
Answer: The provided differential equation is \( (1 + x^2) \frac{dy}{dx} + 2xy = \frac{1}{1+x^2} \). To convert this into the standard linear form \( \frac{dy}{dx} + Py = Q \), we need to divide the entire equation by \( (1 + x^2) \). Dividing by \( (1 + x^2) \), we get: \( \frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{1}{(1 + x^2)(1 + x^2)} \) \( \frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{1}{(1 + x^2)^2} \) Now, we can clearly identify \( P = \frac{2x}{1 + x^2} \) and \( Q = \frac{1}{(1 + x^2)^2} \). The next step is to calculate the integrating factor (I.F.). The formula is \( I.F. = e^{\int P dx} \). Substituting the value of \( P \), we get: \( I.F. = e^{\int \frac{2x}{1 + x^2} dx} \) To solve the integral \( \int \frac{2x}{1 + x^2} dx \), we use substitution. Let \( t = 1 + x^2 \). Then, \( dt = 2x dx \). So, the integral becomes \( \int \frac{1}{t} dt = \log |t| \). Substituting back \( t = 1 + x^2 \), we get \( \log (1 + x^2) \) (since \( 1 + x^2 \) is always positive). Therefore, the integrating factor is: \( I.F. = e^{\log (1 + x^2)} = 1 + x^2 \) Next, we multiply both sides of the modified differential equation by this integrating factor \( (1 + x^2) \): \( (1 + x^2) \frac{dy}{dx} + (1 + x^2) \cdot \frac{2x}{1 + x^2} y = (1 + x^2) \cdot \frac{1}{(1 + x^2)^2} \) This simplifies to: \( (1 + x^2) \frac{dy}{dx} + 2xy = \frac{1}{1 + x^2} \) The left side of this equation is the derivative of the product \( y (1 + x^2) \) with respect to \( x \). So, we can write: \( \frac{d}{dx} (y (1 + x^2)) = \frac{1}{1 + x^2} \) To find the general solution for \( y \), we integrate both sides of this equation with respect to \( x \): \( y (1 + x^2) = \int \frac{1}{1 + x^2} dx \) The integral of \( \frac{1}{1 + x^2} \) is \( \tan^{-1} x \). We also include the constant of integration, \( C \). \( y (1 + x^2) = \tan^{-1} x + C \) To find the particular solution, we use the given condition: when \( x = 1 \), \( y = 0 \). Substitute these values into the general solution: \( 0 \cdot (1 + 1^2) = \tan^{-1} (1) + C \) \( 0 \cdot (2) = \frac{\pi}{4} + C \) \( 0 = \frac{\pi}{4} + C \) \( C = -\frac{\pi}{4} \) Now, substitute the value of \( C \) back into the general solution to get the particular solution: \( y (1 + x^2) = \tan^{-1} x - \frac{\pi}{4} \) Finally, to get the expression for \( y \), we divide the entire equation by \( (1 + x^2) \): \( y = \frac{\tan^{-1} x - \frac{\pi}{4}}{1 + x^2} \) This is the particular solution for the given differential equation under the specified condition.
In simple words: First, we change the equation to the standard linear form. Then, we find the integrating factor using substitution. We multiply the equation by this factor, recognize the left side as a product's derivative, and integrate both sides. Finally, we use the given conditions to find the constant \( C \) and write the particular solution.

Exam Tip: Ensure that the initial transformation to the standard linear form is correct. Remember the integral of \( \frac{1}{1+x^2} \) as \( \tan^{-1} x \). Substitute the initial conditions carefully to calculate the constant \( C \).

Question 14. \( (1 + x^{2})\frac{d y}{d x} + 2xy = \frac{1}{1 +x^2} \) when \( x = 1 \), then \( y = 0 \). Find the particular solution of the differential equation.
Answer: The given differential equation is \( (1 + x^{2})\frac{d y}{d x} + 2xy = \frac{1}{1 +x^2} \).
We first divide the equation by \( (1 + x^{2}) \) to put it in the standard form \( \frac{d y}{d x} + Py = Q \).
So, \( \frac{d y}{d x} + \frac{2x}{1+x^{2}} y = \frac{1}{(1+x^{2})^2} \) (1)
Comparing this with \( \frac{d y}{d x} + Py = Q \), we get \( P = \frac{2x}{1+x^{2}} \) and \( Q = \frac{1}{(1+x^{2})^2} \).
The integrating factor (I.F.) is calculated as \( e^{\int P \, dx} \).
\( I.F. = e^{\int \frac{2x}{1+x^{2}} \, dx} \)
To solve the integral, let \( t = 1+x^{2} \), so \( dt = 2x \, dx \).
\( \int \frac{2x}{1+x^{2}} \, dx = \int \frac{1}{t} \, dt = \log |t| = \log |1+x^{2}| \).
Therefore, \( I.F. = e^{\log |1+x^{2}|} = 1+x^{2} \).
Now, we multiply both sides of equation (1) by the integrating factor \( (1+x^{2}) \).
\( (1+x^{2})\frac{d y}{d x} + (1+x^{2})\frac{2x}{1+x^{2}} y = (1+x^{2})\frac{1}{(1+x^{2})^2} \)
\( (1+x^{2})\frac{d y}{d x} + 2xy = \frac{1}{1+x^{2}} \)
The left side is the derivative of \( y \cdot (I.F.) \).
\( \frac{d}{d x}(y(1+x^{2})) = \frac{1}{1+x^{2}} \)
Next, we integrate both sides with respect to \( x \).
\( \int \frac{d}{d x}(y(1+x^{2})) \, dx = \int \frac{1}{1+x^{2}} \, dx \)
\( y(1+x^{2}) = \tan^{-1}x + c \)
This is the general solution. Now, we use the given condition that when \( x = 1 \), then \( y = 0 \), to find the value of \( c \).
\( 0(1+1^{2}) = \tan^{-1}(1) + c \)
\( 0 = \frac{\pi}{4} + c \)
\( \implies c = -\frac{\pi}{4} \)
Substituting the value of \( c \) back into the general solution, we get the particular solution:
\( y(1+x^{2}) = \tan^{-1}x - \frac{\pi}{4} \).
In simple words: We changed the equation into a standard form, found a special multiplier called the integrating factor, and then used it to solve the equation. Finally, we put in the given \( x \) and \( y \) values to find the specific constant for this problem.

Exam Tip: Remember to always rearrange the differential equation into the standard linear form \( \frac{dy}{dx} + Py = Q \) (or \( \frac{dx}{dy} + P_1x = Q_1 \)) before identifying P and Q. Also, carefully apply the given initial conditions to find the particular solution's constant.

Questions 13 To 15: Find The Particular Solutions Of The Following Differential Equations Under The Given Conditions.

Question 15. \( \frac{d y}{d x} – 3y \cot x = \sin 2x \), when \( x = \frac{\pi}{2} \) then \( y = 2 \).
Answer: The given differential equation is \( \frac{d y}{d x} – 3y \cot x = \sin 2x \) (1)
This equation is already in the standard linear form \( \frac{d y}{d x} + Py = Q \).
Comparing, we have \( P = -3 \cot x \) and \( Q = \sin 2x \).
Next, we find the integrating factor (I.F.):
\( I.F. = e^{\int P \, dx} = e^{\int -3 \cot x \, dx} \)
\( \int -3 \cot x \, dx = -3 \int \frac{\cos x}{\sin x} \, dx = -3 \log |\sin x| = \log (|\sin x|^{-3}) \)
So, \( I.F. = e^{\log (|\sin x|^{-3})} = |\sin x|^{-3} = \frac{1}{\sin^3 x} \).
Now, we multiply equation (1) by the integrating factor:
\( \frac{1}{\sin^3 x} \frac{d y}{d x} - \frac{3 \cot x}{\sin^3 x} y = \frac{\sin 2x}{\sin^3 x} \)
The left side is \( \frac{d}{d x}(y \cdot I.F.) \).
\( \frac{d}{d x}\left(y \cdot \frac{1}{\sin^3 x}\right) = \frac{\sin 2x}{\sin^3 x} \)
We know \( \sin 2x = 2 \sin x \cos x \).
\( \frac{d}{d x}\left(y \cdot \frac{1}{\sin^3 x}\right) = \frac{2 \sin x \cos x}{\sin^3 x} = \frac{2 \cos x}{\sin^2 x} = 2 \cot x \csc x \)
Now, integrate both sides with respect to \( x \):
\( \int \frac{d}{d x}\left(y \cdot \frac{1}{\sin^3 x}\right) \, dx = \int 2 \cot x \csc x \, dx \)
\( y \cdot \frac{1}{\sin^3 x} = -2 \csc x + c \)
This is the general solution. Now, we use the given condition: when \( x = \frac{\pi}{2} \), then \( y = 2 \).
\( 2 \cdot \frac{1}{\sin^3 \left(\frac{\pi}{2}\right)} = -2 \csc \left(\frac{\pi}{2}\right) + c \)
Since \( \sin \left(\frac{\pi}{2}\right) = 1 \) and \( \csc \left(\frac{\pi}{2}\right) = 1 \):
\( 2 \cdot \frac{1}{1^3} = -2 \cdot 1 + c \)
\( 2 = -2 + c \)
\( \implies c = 4 \)
Substitute \( c = 4 \) back into the general solution:
\( y \cdot \frac{1}{\sin^3 x} = -2 \csc x + 4 \)
Multiply by \( \sin^3 x \):
\( y = -2 \csc x \sin^3 x + 4 \sin^3 x \)
Since \( \csc x = \frac{1}{\sin x} \):
\( y = -2 \frac{1}{\sin x} \sin^3 x + 4 \sin^3 x \)
\( y = -2 \sin^2 x + 4 \sin^3 x \).
This is the particular solution.
In simple words: We solved the differential equation by finding an integrating factor, which helped us to combine the left side into a single derivative. After integrating, we used the given point \( (x, y) \) to find the exact value of the constant, giving us the specific solution for the curve.

Exam Tip: For particular solutions, always solve for the general solution first. Then, substitute the given boundary conditions (x and y values) into the general solution to determine the constant of integration. Make sure to perform all trigonometric simplifications correctly.

Question 16. If the slope of the tangent at any point (x, y) on a curve is equal to the sum of the coordinates of that point, then find the equation of the curve passing through the origin.
Answer: The slope of the tangent at any point \( (x, y) \) on a curve is given by \( \frac{d y}{d x} \).
According to the problem statement, the slope of the tangent is equal to the sum of the coordinates of that point.
So, we can write the differential equation as:
\( \frac{d y}{d x} = x + y \)
Rearranging this into the standard linear differential equation form \( \frac{d y}{d x} + Py = Q \):
\( \frac{d y}{d x} - y = x \) (1)
Here, \( P = -1 \) and \( Q = x \).
Next, we calculate the integrating factor (I.F.):
\( I.F. = e^{\int P \, dx} = e^{\int -1 \, dx} = e^{-x} \).
Now, we multiply both sides of equation (1) by the integrating factor \( e^{-x} \):
\( e^{-x}\frac{d y}{d x} - e^{-x}y = x e^{-x} \)
The left side of this equation can be expressed as the derivative of the product \( y \cdot (I.F.) \).
\( \frac{d}{d x}(y e^{-x}) = x e^{-x} \)
Now, we integrate both sides with respect to \( x \):
\( \int \frac{d}{d x}(y e^{-x}) \, dx = \int x e^{-x} \, dx \)
To solve \( \int x e^{-x} \, dx \), we use integration by parts, \( \int u \, dv = uv - \int v \, du \).
Let \( u = x \) and \( dv = e^{-x} \, dx \).
Then \( du = dx \) and \( v = -e^{-x} \).
\( \int x e^{-x} \, dx = x(-e^{-x}) - \int (-e^{-x}) \, dx \)
\( = -x e^{-x} + \int e^{-x} \, dx \)
\( = -x e^{-x} - e^{-x} + C \)
So, the general solution is:
\( y e^{-x} = -x e^{-x} - e^{-x} + C \)
Divide by \( e^{-x} \) to solve for \( y \):
\( y = -x - 1 + C e^{x} \).
The problem states that the curve passes through the origin, which means it passes through the point \( (0, 0) \).
Substitute \( x = 0 \) and \( y = 0 \) into the general solution to find \( C \):
\( 0 = -0 - 1 + C e^{0} \)
\( 0 = -1 + C \cdot 1 \)
\( \implies C = 1 \).
Substitute \( C = 1 \) back into the general solution to get the equation of the curve:
\( y = -x - 1 + 1 \cdot e^{x} \)
\( y = e^{x} - x - 1 \).
This can also be written as \( y + x + 1 = e^{x} \).
In simple words: We took the problem's description about the slope and set up a first-order differential equation. We then used a method called integrating factor to solve it and found a general formula for the curve. Since the curve goes through the origin \( (0,0) \), we used that point to find the exact constant for our formula.

Exam Tip: Translating word problems into differential equations is a key skill. Clearly define variables, set up the equation based on the given relationships, and then solve using the appropriate method (like integrating factor for linear first-order ODEs). Always use the initial condition to find the particular solution.

Question 17. If the sum of the coordinates of any point (x, y) on a curve is 5 more than the value of the slope of the tangent at that point, then find the equation of the curve passing through the point (0, 2).
Answer: The slope of the tangent at any point \( (x, y) \) on a curve is given by \( \frac{d y}{d x} \).
According to the problem statement, the sum of the coordinates \( (x + y) \) is 5 more than the slope of the tangent.
So, we can write the differential equation as:
\( x + y = \frac{d y}{d x} + 5 \)
Rearranging this into the standard linear differential equation form \( \frac{d y}{d x} + Py = Q \):
\( \frac{d y}{d x} - y = x - 5 \) (1)
Here, \( P = -1 \) and \( Q = x - 5 \).
Next, we calculate the integrating factor (I.F.):
\( I.F. = e^{\int P \, dx} = e^{\int -1 \, dx} = e^{-x} \).
Now, we multiply both sides of equation (1) by the integrating factor \( e^{-x} \):
\( e^{-x}\frac{d y}{d x} - e^{-x}y = (x - 5) e^{-x} \)
The left side of this equation can be expressed as the derivative of the product \( y \cdot (I.F.) \).
\( \frac{d}{d x}(y e^{-x}) = (x - 5) e^{-x} \)
Now, we integrate both sides with respect to \( x \):
\( \int \frac{d}{d x}(y e^{-x}) \, dx = \int (x - 5) e^{-x} \, dx \)
To solve \( \int (x - 5) e^{-x} \, dx \), we use integration by parts, \( \int u \, dv = uv - \int v \, du \).
Let \( u = x - 5 \) and \( dv = e^{-x} \, dx \).
Then \( du = dx \) and \( v = -e^{-x} \).
\( \int (x - 5) e^{-x} \, dx = (x - 5)(-e^{-x}) - \int (-e^{-x}) \, dx \)
\( = -(x - 5)e^{-x} + \int e^{-x} \, dx \)
\( = -(x - 5)e^{-x} - e^{-x} + C \)
So, the general solution is:
\( y e^{-x} = -(x - 5)e^{-x} - e^{-x} + C \)
Divide by \( e^{-x} \) to solve for \( y \):
\( y = -(x - 5) - 1 + C e^{x} \)
\( y = -x + 5 - 1 + C e^{x} \)
\( y = -x + 4 + C e^{x} \).
The problem states that the curve passes through the point \( (0, 2) \).
Substitute \( x = 0 \) and \( y = 2 \) into the general solution to find \( C \):
\( 2 = -0 + 4 + C e^{0} \)
\( 2 = 4 + C \cdot 1 \)
\( \implies C = -2 \).
Substitute \( C = -2 \) back into the general solution to get the equation of the curve:
\( y = -x + 4 - 2 e^{x} \).
In simple words: We formed a differential equation from the given information about the slope and coordinates. We then found the integrating factor to solve it and derive the general form of the curve. Using the specified point \( (0,2) \) that the curve passes through, we determined the exact constant, giving us the specific curve's equation.

Exam Tip: Be careful with signs when setting up the differential equation from the word problem. Ensure all terms are correctly moved to fit the standard linear form. When integrating with parts, double-check your 'u' and 'dv' assignments and their derivatives/integrals to prevent errors.

Questions 18 And 19: Choose The Correct Option To Make The Given Statement True.

Question 18. The integrating factor of the differential equation \( x\frac{d y}{d x} – y = 2x^2 \) is _______.
(A) \( e^{-x} \)
(B) \( e^{-y} \)
(C) \( \frac{1}{x} \)
(D) \( x \)
Answer: (C) \( \frac{1}{x} \)
In simple words: To find the integrating factor, first rewrite the equation to a standard form, then find the value of P and use the formula \( e^{\int P \, dx} \). This calculation leads to \( \frac{1}{x} \).

Exam Tip: For MCQs on integrating factors, first convert the given differential equation into the standard linear form \( \frac{dy}{dx} + Py = Q \). Then, correctly identify P and compute \( e^{\int P \, dx} \). Watch for simple algebraic mistakes, especially with signs or division by variables like x.

Question 19. The integrating factor of the differential equation \( (1 – y^{2})\frac{d x}{d y} + yx = ay (-1 < y < 1) \) is _______.
(A) \( \frac{1}{y^2-1} \)
(B) \( \frac{1}{\sqrt{y^2-1}} \)
(C) \( \frac{1}{1-y^2} \)
(D) \( \frac{1}{\sqrt{1-y^2}} \)
Answer: (D) \( \frac{1}{\sqrt{1-y^2}} \)
In simple words: We need to get the equation into a standard form for \( \frac{dx}{dy} \), identify the \( P_1 \) term, and then calculate \( e^{\int P_1 \, dy} \). This will give us \( \frac{1}{\sqrt{1-y^2}} \).

Exam Tip: When the differential equation is in terms of \( \frac{dx}{dy} \), remember that the standard form is \( \frac{dx}{dy} + P_1x = Q_1 \) and the integrating factor is \( e^{\int P_1 \, dy} \). Pay close attention to the variable of integration (dy instead of dx) and the bounds or conditions given for y.