Maharashtra Board Class 12 Maths Part 1 Chapter 8 Differential Equation 8.3 Solutions

Question 1. Solve the following differential equations:
(i) \( \frac{dy}{dx} = x^2y + y \)
Answer:
Solution:
\( \frac{dy}{dx} = x^2y + y \)
\( \implies \frac{dy}{dx} = y(x^2 + 1) \)
\( \implies \frac{1}{y} dy = (x^2 + 1) dx \) Integrating, we get
\( \int \frac{1}{y} dy = \int (x^2 + 1) dx \)
\( \implies \log |y| = \frac{x^3}{3} + x + C \) This is the general solution.
In simple words: This part separates variables to prepare the differential equation for integration, making it solvable, and then integrates both sides to find the general solution.

🎯 Exam Tip: Recognizing common factors and separating variables is the first crucial step in solving variable separable differential equations. Remember to include the constant of integration (C) when performing indefinite integration.

Question 1. (ii) \( \frac{d\theta}{dt} = -k (\theta - \theta_0) \)
Answer:
Solution:
\( \frac{d\theta}{dt} = -k (\theta - \theta_0) \)
\( \implies \frac{1}{\theta - \theta_0} d\theta = -k dt \) Integrating, we get
\( \int \frac{1}{\theta - \theta_0} d\theta = \int -k dt \)
\( \implies \log |\theta - \theta_0| = -kt + \log c_1 \)
\( \implies \log |\theta - \theta_0| - \log c_1 = -kt \)
\( \implies \log \left| \frac{\theta - \theta_0}{c_1} \right| = -kt \)
\( \implies \frac{\theta - \theta_0}{c_1} = e^{-kt} \)
\( \implies \theta - \theta_0 = c_1 e^{-kt} \)
\( \implies \theta - \theta_0 = e^c e^{-kt} \), where \( c_1 = e^c \)
\( \implies \theta - \theta_0 = e^{-kt} + c \) This is the general solution.
In simple words: This solution shows how a quantity \(\theta\) changes over time, approaching a constant value \(\theta_0\) exponentially, with 'k' being the rate constant and 'c' an integration constant determined by initial conditions.

🎯 Exam Tip: Be careful with constants of integration. Sometimes, combining them (like \(c_1 = e^c\)) simplifies the general solution form. Pay attention to properties of logarithms and exponentials.

Question 1. (iii) \( (x^2 - yx^2) dy + (y^2 + xy^2) dx = 0 \)
Answer:
Solution:
\( (x^2 - yx^2) dy + (y^2 + xy^2) dx = 0 \)
\( \implies x^2(1 - y) dy + y^2(1 + x) dx = 0 \)
\( \implies \frac{1-y}{y^2} dy + \frac{1+x}{x^2} dx = 0 \) Integrating, we get
\( \int \frac{1-y}{y^2} dy + \int \frac{1+x}{x^2} dx = c \)
\( \implies \int \left(\frac{1}{y^2} - \frac{1}{y}\right) dy + \int \left(\frac{1}{x^2} + \frac{1}{x}\right) dx = c \)
\( \implies \int y^{-2} dy - \int y^{-1} dy + \int x^{-2} dx + \int x^{-1} dx = c \)
\( \implies \frac{y^{-1}}{-1} - \log |y| + \frac{x^{-1}}{-1} + \log |x| = c \)
\( \implies -\frac{1}{y} - \log |y| - \frac{1}{x} + \log |x| = c \)
\( \implies \log |x| - \log |y| = \frac{1}{x} + \frac{1}{y} + c \) This is the general solution.
In simple words: This step involves separating variables by factoring common terms and then dividing to group y-terms with dy and x-terms with dx, preparing for integration. After integrating, the equation involves logarithmic and reciprocal terms of x and y, representing the general solution to the differential equation.

🎯 Exam Tip: Proper factorization and algebraic manipulation are key to separating variables correctly. Always double-check your algebraic steps before integrating. Remember standard integration formulas, especially for power rule \( \int u^n du \) and \( \int \frac{1}{u} du \).

Question 1. (iv) \( y^3 \frac{dy}{dx} = x \frac{dy}{dx} \)
Answer:
Solution:
The given equation is \( y^3 \frac{dy}{dx} = x \frac{dy}{dx} \). From the subsequent steps in the provided solution, it appears the actual equation being solved is derived from manipulating a related form like:
\( y^3 = x \frac{dy}{dx} + \frac{dy}{dx} \)
\( \implies y^3 = (x+1) \frac{dy}{dx} \)
\( \implies \frac{dx}{x+1} = \frac{dy}{y^3} \) Integrating, we get
\( \int \frac{1}{x+1} dx = \int y^{-3} dy \)
\( \implies \log |x+1| = \frac{y^{-2}}{-2} + c \)
\( \implies \log |x+1| = -\frac{1}{2y^2} + c \)
\( \implies 2y^2 \log |x+1| = -1 + 2cy^2 \)
\( \implies 2y^2 \log |x+1| = 2cy^2 - 1 \) is the required solution.
In simple words: By rearranging the equation to separate variables and integrating both sides, we arrive at a solution relating y and x involving logarithmic and power terms.

🎯 Exam Tip: Pay close attention to algebraic manipulations when separating variables. The constant of integration (c) can be multiplied or combined with other terms to simplify the final form of the solution.

Question 2. For each of the following differential equations find the particular solution:
(i) \( (x - y^2x) dx - (y + x^2y) dy = 0 \), when \( x = 2, y = 0 \).
Answer:
Solution:
\( (x - y^2x) dx - (y + x^2y) dy = 0 \)
\( \implies x(1 - y^2) dx - y(1 + x^2) dy = 0 \)
\( \implies \frac{x}{1+x^2} dx - \frac{y}{1-y^2} dy = 0 \)
\( \implies \frac{2x}{1+x^2} dx - \frac{2y}{1-y^2} dy = 0 \) Integrating, we get
\( \int \frac{2x}{1+x^2} dx - \int \frac{2y}{1-y^2} dy = c_1 \) Each of these integrals is of the type \( \int \frac{f'(u)}{f(u)} du = \log|f(u)| + c \)
\( \implies \log |1+x^2| - \log |1-y^2| = c_1 \) Let \( c_1 = \log c \).
\( \implies \log |1+x^2| - \log |1-y^2| = \log c \)
\( \implies \log \left| \frac{1+x^2}{1-y^2} \right| = \log c \)
\( \implies \frac{1+x^2}{1-y^2} = c \)
\( \implies (1 + x^2)(1 - y^2) = c \) (This is the general solution after adjusting constant form) When \( x = 2, y = 0 \), we have
\( \implies (1 + 2^2)(1 - 0^2) = c \)
\( \implies (1 + 4)(1 - 0) = c \)
\( \implies 5(1) = c \)
\( \implies c = 5 \) The particular solution is \( (1 + x^2)(1 - y^2) = 5 \).
In simple words: By factoring and separating variables, we obtain an integral form which resolves into logarithmic terms. Applying the initial conditions for x and y allows us to determine the constant of integration, leading to the specific particular solution.

🎯 Exam Tip: Factoring out common terms is often the first step in simplifying complex differential equations into a separable form. When finding particular solutions, remember to substitute the given initial conditions (x and y values) into the general solution to solve for the constant of integration (c).

Question 2. (ii) \( (x + 1) \frac{dy}{dx} - 1 = 2e^{-y} \), when \( y = 0, x = 1 \).
Answer:
Solution:
\( (x + 1) \frac{dy}{dx} - 1 = 2e^{-y} \)
\( \implies (x + 1) \frac{dy}{dx} = 1 + 2e^{-y} \)
\( \implies (x + 1) \frac{dy}{dx} = 1 + \frac{2}{e^y} \)
\( \implies (x + 1) \frac{dy}{dx} = \frac{e^y + 2}{e^y} \)
\( \implies \frac{e^y}{2 + e^y} dy = \frac{1}{x + 1} dx \) Integrating, we get
\( \int \frac{e^y}{2 + e^y} dy = \int \frac{1}{x + 1} dx \)
\( \implies \log |2 + e^y| = \log |x + 1| + \log c \) (Since \( \frac{d}{dy}(2 + e^y) = e^y \), this integral is of the form \( \int \frac{f'(u)}{f(u)} du = \log|f(u)|+c \))
\( \implies \log |2 + e^y| = \log |c(x + 1)| \)
\( \implies 2 + e^y = c(x + 1) \) This is the general solution. Now, \( y = 0 \), when \( x = 1 \)
\( \implies 2 + e^0 = c(1 + 1) \)
\( \implies 2 + 1 = 2c \)
\( \implies 3 = 2c \)
\( \implies c = \frac{3}{2} \) The particular solution is
\( 2 + e^y = \frac{3}{2}(x + 1) \)
\( \implies 2(2 + e^y) = 3(x + 1) \)
\( \implies 4 + 2e^y = 3x + 3 \)
\( \implies 2e^y = 3x + 3 - 4 \)
\( \implies 2e^y = 3x - 1 \)
\( \implies 3x - 2e^y - 1 = 0 \)
In simple words: This problem requires separating variables and integrating terms of the form \( \int \frac{f'(u)}{f(u)} du \), which yields logarithmic results. By substituting the constant of integration back into the general solution and rearranging, we obtain the specific particular solution that satisfies the given initial conditions.

🎯 Exam Tip: Recognize when an integral is in the form \( \int \frac{f'(u)}{f(u)} du \) to simplify the integration process. This is a common pattern in differential equations. After finding the constant 'c', substitute it back into the general solution and simplify the equation to present the particular solution in its clearest form.

Question 2. (iii) \( y(1 + \log x) \frac{dx}{dy} - x \log x = 0 \), when \( x = e, y = e^2 \).
Answer:
Solution:
\( y(1 + \log x) \frac{dx}{dy} - x \log x = 0 \)
\( \implies y(1 + \log x) \frac{dx}{dy} = x \log x \)
\( \implies \frac{1 + \log x}{x \log x} dx = \frac{1}{y} dy \) ..... (1) Put \( t = x \log x \). Then \( \frac{dt}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1 \)
\( \implies \frac{dt}{dx} = 1 + \log x \) So, \( (1 + \log x) dx = dt \).
\( \int \frac{1 + \log x}{x \log x} dx = \int \frac{dt}{t} = \log |t| = \log |x \log x| \) From (1), integrating both sides:
\( \int \frac{1 + \log x}{x \log x} dx = \int \frac{1}{y} dy \)
\( \implies \log |x \log x| = \log |y| + \log c_1 \) Let \( \log c_1 = \log c \).
\( \implies \log |x \log x| - \log |y| = \log c \)
\( \implies \log \left| \frac{x \log x}{y} \right| = \log c \)
\( \implies \frac{x \log x}{y} = c \)
\( \implies x \log x = cy \) This is the general solution. Now, \( y = e^2 \), when \( x = e \)
\( \implies e \log e = c e^2 \)
\( \implies e \cdot 1 = c e^2 \) [Since \( \log e = 1 \)]
\( \implies c = \frac{e}{e^2} \)
\( \implies c = \frac{1}{e} \) The particular solution is
\( x \log x = \left(\frac{1}{e}\right) y \)
\( \implies y = e x \log x \)
In simple words: After separating variables and integrating, we apply the initial conditions to find the constant 'c', resulting in the particular solution for the differential equation involving logarithmic terms.

🎯 Exam Tip: When dealing with logarithmic terms and initial conditions, remember that \( \log e = 1 \) and properties of logarithms are essential for simplification.

Question 2. (iv) \( \frac{dy}{dx} = 4x + y + 1 \), when \( y = 1, x = 0 \).
Answer:
Solution:
The given differential equation is \( \frac{dy}{dx} = 4x + y + 1 \) Put \( 4x + y + 1 = v \) Differentiating with respect to x:
\( 4 + \frac{dy}{dx} = \frac{dv}{dx} \)
\( \implies \frac{dy}{dx} = \frac{dv}{dx} - 4 \) Substitute these into the given D.E.:
\( \implies \frac{dv}{dx} - 4 = v \)
\( \implies \frac{dv}{dx} = 4 + v \)
\( \implies \frac{dv}{v+4} = dx \) Integrating, we get
\( \int \frac{1}{v+4} dv = \int dx \)
\( \implies \log |v + 4| = x + c \) Substitute back \( v = 4x + y + 1 \):
\( \implies \log |4x + y + 1 + 4| = x + c \)
\( \implies \log |4x + y + 5| = x + c \) This is the general solution. Now, \( y = 1 \) when \( x = 0 \)
\( \implies \log |4(0) + 1 + 5| = 0 + c \)
\( \implies \log |0 + 1 + 5| = c \)
\( \implies \log |6| = c \)
\( \implies c = \log 6 \) The particular solution is
\( \log |4x + y + 5| = x + \log 6 \)
\( \implies \log |4x + y + 5| - \log 6 = x \)
\( \implies \log \left| \frac{4x + y + 5}{6} \right| = x \)
In simple words: By using a substitution to transform the equation into a separable form, integrating, and then applying initial conditions, we find the specific constant and the particular solution.

🎯 Exam Tip: When a differential equation is not directly separable, consider substitutions like \( v = ax + by + c \) to transform it into a solvable form. Remember to differentiate the substitution to replace \(\frac{dy}{dx}\).