Maharashtra Board Class 12 Maths Part 1 Chapter 8 Differential Equation 8.4 Solutions

Solve The Following Differential Equations:

Question 1. x dx + 2y dy = 0
Answer:
x dx + 2y dy = 0
Integrating, we get
\( \int x \, dx + 2 \int y \, dy = C_1 \)
\( \frac{x^2}{2} + 2 \left( \frac{y^2}{2} \right) = C_1 \)
\( \implies x^2 + 2y^2 = c \), where \( c = 2C_1 \)
This is the general solution.
In simple words: To solve this differential equation, we integrate both sides with respect to their respective variables and combine the constants of integration. The result is a simple equation relating x and y.

🎯 Exam Tip: Remember to include the constant of integration when performing indefinite integrals, and simplify the constant where possible to present a clean general solution.

Question 2. y² dx + (xy + x²) dy = 0
Answer:
y² dx + (xy + x²) dy = 0
\( \implies (xy + x^2) dy = -y^2 dx \)
\( \implies \frac{dy}{dx} = \frac{-y^2}{xy+x^2} \) ......(1)
Put \( y = vx \)
\( \implies \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Substituting these values in (1), we get
\( v+x\frac{dv}{dx} = \frac{-v^2x^2}{x \cdot vx+x^2} = \frac{-v^2}{v+1} \)
\( \implies x\frac{dv}{dx} = \frac{-v^2}{v+1} - v \)
\( \implies x\frac{dv}{dx} = \frac{-v^2 - v^2 - v}{v+1} \)
\( \implies x\frac{dv}{dx} = \frac{-2v^2-v}{v+1} = \frac{-(2v^2+v)}{v+1} \)
\( \implies \frac{v+1}{2v^2+v} dv = -\frac{1}{x} dx \)
Integrating, we get
\( \int \frac{v+1}{2v^2+v} dv = \int -\frac{1}{x} dx \)
\( \implies \int \frac{v+1}{v(2v+1)} dv = -\int \frac{1}{x} dx \)
\( \implies \int \left( \frac{1}{v} - \frac{1}{2v+1} \right) dv = -\int \frac{1}{x} dx \)
\( \implies \int \frac{1}{v} dv - \int \frac{1}{2v+1} dv = -\int \frac{1}{x} dx \)
\( \implies \log|v| - \frac{1}{2} \log|2v+1| = -\log|x|+\log c \)
\( \implies 2\log|v| - \log|2v+1| = -2\log|x|+2\log c \)
\( \implies \log|v^2| - \log|2v+1| = -\log|x^2| + \log c^2 \)
\( \implies \log \left| \frac{v^2}{2v+1} \right| = \log \left| \frac{c^2}{x^2} \right| \)
\( \implies \frac{v^2}{2v+1} = \frac{c^2}{x^2} \)
Substitute \( v = \frac{y}{x} \):
\( \implies \frac{(y/x)^2}{2(y/x)+1} = \frac{c^2}{x^2} \)
\( \implies \frac{y^2/x^2}{(2y+x)/x} = \frac{c^2}{x^2} \)
\( \implies \frac{y^2}{x(2y+x)} = \frac{c^2}{x^2} \)
\( \implies y^2x = c^2 (2y+x) \)
\( \implies xy^2 = c^2 (x + 2y) \)
This is the general solution.
In simple words: This is a homogeneous differential equation, solved by substituting \( y=vx \). After separation of variables and integration, we substitute back \( v=y/x \) and simplify to get the general solution in terms of x and y.

🎯 Exam Tip: For homogeneous equations, the substitution \( y=vx \) is key. Be careful with algebraic manipulations during the separation of variables and simplify the logarithmic terms using log properties before substituting back.

Question 3. x²y dx - (x³ + y³) dy = 0
Answer:
x²y dx - (x³ + y³) dy = 0
\( \implies (x^3 + y^3) dy = x^2y dx \)
\( \implies \frac{dy}{dx} = \frac{x^2y}{x^3+y^3} \) ......(1)
Put \( y = vx \)
\( \implies \frac{dy}{dx} = v + x \frac{dv}{dx} \)
\( \implies \) (1) becomes,
\( v+x\frac{dv}{dx} = \frac{x^2 \cdot vx}{x^3+v^3x^3} = \frac{vx^3}{x^3(1+v^3)} = \frac{v}{1+v^3} \)
\( \implies x\frac{dv}{dx} = \frac{v}{1+v^3} - v \)
\( \implies x\frac{dv}{dx} = \frac{v - v(1+v^3)}{1+v^3} = \frac{v - v - v^4}{1+v^3} \)
\( \implies x\frac{dv}{dx} = \frac{-v^4}{1+v^3} \)
\( \implies \frac{1+v^3}{v^4} dv = -\frac{1}{x} dx \)
Integrating, we get
\( \int \frac{1+v^3}{v^4} dv = \int -\frac{1}{x} dx \)
\( \implies \int \left( v^{-4} + \frac{1}{v} \right) dv = -\int \frac{1}{x} dx \)
\( \implies \int v^{-4} dv + \int \frac{1}{v} dv = -\int \frac{1}{x} dx \)
\( \implies \frac{v^{-3}}{-3} + \log|v| = -\log|x|+C_1 \)
\( \implies -\frac{1}{3v^3} + \log|v| = -\log|x|+C_1 \)
Substitute \( v = \frac{y}{x} \):
\( \implies -\frac{1}{3(y/x)^3} + \log\left|\frac{y}{x}\right| = -\log|x|+C_1 \)
\( \implies -\frac{x^3}{3y^3} + \log|y| - \log|x| = -\log|x|+C_1 \)
\( \implies -\frac{x^3}{3y^3} + \log|y| = C_1 \)
Let \( C_1 = -\log c \)
\( \implies -\frac{x^3}{3y^3} = -\log|y| - \log c \)
\( \implies \frac{x^3}{3y^3} = \log|y| + \log c \)
\( \implies \frac{x^3}{3y^3} = \log|cy| \)
This is the general solution.
In simple words: This is a homogeneous differential equation. We use the substitution \( y=vx \), separate the variables, and then integrate. After substituting back for \( v \), we rearrange the terms to find the general solution.

🎯 Exam Tip: When integrating terms like \( \frac{1+v^3}{v^4} \), split them into simpler fractions (\( v^{-4} + v^{-1} \)) to integrate easily. Be careful with signs and logarithmic properties during simplification.

Question 4. \( \frac{dy}{dx} + \frac{x-2y}{2x-y} = 0 \)
Answer:
\( \frac{dy}{dx} = -\frac{x-2y}{2x-y} = \frac{2y-x}{2x-y} \) .........(1)
Put \( y = vx \)
\( \implies \frac{dy}{dx} = v + x \frac{dv}{dx} \)
\( \implies \) (1) becomes,
\( v+x\frac{dv}{dx} = \frac{2vx-x}{2x-vx} = \frac{x(2v-1)}{x(2-v)} = \frac{2v-1}{2-v} \)
\( \implies x\frac{dv}{dx} = \frac{2v-1}{2-v} - v \)
\( \implies x\frac{dv}{dx} = \frac{2v-1 - v(2-v)}{2-v} = \frac{2v-1 - 2v + v^2}{2-v} \)
\( \implies x\frac{dv}{dx} = \frac{v^2-1}{2-v} \)
\( \implies \frac{2-v}{v^2-1} dv = \frac{1}{x} dx \)
Integrating, we get
\( \int \frac{2-v}{v^2-1} dv = \int \frac{1}{x} dx \)
\( \implies \int \frac{2}{v^2-1} dv - \int \frac{v}{v^2-1} dv = \int \frac{1}{x} dx \)
\( \implies 2 \int \frac{1}{v^2-1} dv - \frac{1}{2} \int \frac{2v}{v^2-1} dv = \int \frac{1}{x} dx \)
\( \implies 2 \cdot \frac{1}{2} \log \left| \frac{v-1}{v+1} \right| - \frac{1}{2} \log|v^2-1| = \log|x|+\log c_1 \)
\( \left[ \frac{d}{dv} (v^2-1) = 2v \text{ and } \int \frac{f'(x)}{f(x)} \, dx = \log|f(x)|+c \right] \)
\( \implies \log \left| \frac{v-1}{v+1} \right| - \log | (v^2-1)^{1/2} | = \log|c_1x| \)
\( \implies \log \left| \frac{v-1}{v+1} \cdot \frac{1}{\sqrt{v^2-1}} \right| = \log|c_1x| \)
\( \implies \frac{v-1}{v+1} \cdot \frac{1}{\sqrt{v^2-1}} = c_1x \)
\( \implies \frac{v-1}{v+1} \cdot \frac{1}{\sqrt{(v-1)(v+1)}} = c_1x \)
\( \implies \frac{\sqrt{v-1}}{\sqrt{v+1}} = c_1x \)
Substitute \( v = \frac{y}{x} \):
\( \implies \frac{\sqrt{\frac{y}{x}-1}}{\sqrt{\frac{y}{x}+1}} = c_1x \)
\( \implies \frac{\sqrt{\frac{y-x}{x}}}{\sqrt{\frac{y+x}{x}}} = c_1x \)
\( \implies \sqrt{\frac{y-x}{y+x}} = c_1x \)
Squaring both sides:
\( \frac{y-x}{y+x} = c_1^2x^2 \)
Let \( c = c_1^2 \)
\( \implies \frac{y-x}{y+x} = cx^2 \)
This is the general solution.
In simple words: This is a homogeneous equation, so we substitute \( y=vx \). After separating variables and integrating, we use logarithmic properties to simplify the expression. Finally, we substitute back \( v=y/x \) and simplify to get the general solution.

🎯 Exam Tip: When integrating fractions, partial fraction decomposition or recognizing \( \int \frac{f'(x)}{f(x)} \, dx \) forms is crucial. Remember to rationalize or simplify square root terms and combine constants correctly.

Question 5. (x² - y²) dx + 2xy dy = 0
Answer:
(x² - y²) dx + 2xy dy = 0
\( \implies 2xy dy = -(x^2 - y^2) dx = (y^2 - x^2) dx \)
\( \implies \frac{dy}{dx} = \frac{y^2-x^2}{2xy} \) .........(1)
Put \( y = vx \)
\( \implies \frac{dy}{dx} = v+x\frac{dv}{dx} \)
\( \implies \) (1) becomes,
\( v+x\frac{dv}{dx} = \frac{(vx)^2-x^2}{2x(vx)} = \frac{v^2x^2-x^2}{2vx^2} = \frac{x^2(v^2-1)}{2vx^2} = \frac{v^2-1}{2v} \)
\( \implies x\frac{dv}{dx} = \frac{v^2-1}{2v} - v \)
\( \implies x\frac{dv}{dx} = \frac{v^2-1 - 2v^2}{2v} = \frac{-1-v^2}{2v} = -\frac{1+v^2}{2v} \)
\( \implies \frac{2v}{1+v^2} dv = -\frac{1}{x} dx \)
Integrating, we get
\( \int \frac{2v}{1+v^2} dv = \int -\frac{1}{x} dx \)
\( \implies \log|1+v^2| = -\log|x|+\log c \)
\( \left[ \frac{d}{dv} (1+v^2) = 2v \text{ and } \int \frac{f'(x)}{f(x)} \, dx = \log|f(x)|+c \right] \)
\( \implies \log|1+v^2| + \log|x| = \log c \)
\( \implies \log|x(1+v^2)| = \log c \)
\( \implies x(1+v^2) = c \)
Substitute \( v = \frac{y}{x} \):
\( \implies x\left(1+\frac{y^2}{x^2}\right) = c \)
\( \implies x\left(\frac{x^2+y^2}{x^2}\right) = c \)
\( \implies \frac{x^2+y^2}{x} = c \)
\( \implies x^2+y^2 = cx \)
This is the general solution.
In simple words: We solve this homogeneous differential equation by substituting \( y=vx \). This transforms the equation into a separable form. After integration and using logarithmic properties, we substitute \( v=y/x \) back to obtain the final general solution.

🎯 Exam Tip: Recognize the derivative-in-numerator form \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| \) for efficient integration. Algebraic simplification after substituting back \( v=y/x \) is critical for the final solution form.

Question 6. \( xy \frac{dy}{dx} = x^2 + 2y^2 \)
Answer:
\( \frac{dy}{dx} = \frac{x^2+2y^2}{xy} \) ...... (1)
Put \( y = vx \). Then \( \frac{dy}{dx} = v+x\frac{dv}{dx} \)
\( \implies \) (1) becomes,
\( v+x\frac{dv}{dx} = \frac{x^2+2(vx)^2}{x(vx)} = \frac{x^2+2v^2x^2}{vx^2} = \frac{x^2(1+2v^2)}{vx^2} = \frac{1+2v^2}{v} \)
\( \implies x\frac{dv}{dx} = \frac{1+2v^2}{v} - v \)
\( \implies x\frac{dv}{dx} = \frac{1+2v^2-v^2}{v} = \frac{1+v^2}{v} \)
\( \implies \frac{v}{1+v^2} dv = \frac{1}{x} dx \)
Integrating, we get
\( \int \frac{v}{1+v^2} dv = \int \frac{1}{x} dx \)
\( \implies \frac{1}{2} \int \frac{2v}{1+v^2} dv = \int \frac{1}{x} dx + \log c_1 \)
\( \implies \frac{1}{2}\log|1+v^2| = \log|x|+\log c_1 \)
\( \implies \log|1+v^2| = 2\log|x|+2\log c_1 \)
\( \implies \log|1+v^2| = \log|x^2|+\log c_1^2 \)
\( \implies \log|1+v^2| = \log|c_1^2x^2| \)
\( \implies 1+v^2 = c_1^2x^2 \)
Let \( c = c_1^2 \)
\( \implies 1+v^2 = cx^2 \)
Substitute \( v = \frac{y}{x} \):
\( \implies 1+\frac{y^2}{x^2} = cx^2 \)
\( \implies \frac{x^2+y^2}{x^2} = cx^2 \)
\( \implies x^2+y^2 = cx^4 \)
This is the general solution.
In simple words: This is a homogeneous differential equation. We apply the substitution \( y=vx \) to separate variables. After integrating both sides, we use logarithmic properties and substitute \( v=y/x \) back into the equation to find the general solution.

🎯 Exam Tip: When integrating \( \frac{v}{1+v^2} \), multiply and divide by 2 to get the \( \frac{f'(x)}{f(x)} \) form. Remember to combine the constant terms after integrating and express the final solution in terms of x and y.

Question 7. \( x^2 \frac{dy}{dx} = x^2 + xy - y^2 \)
Answer:
\( \frac{dy}{dx} = \frac{x^2+xy-y^2}{x^2} = 1+\frac{y}{x}-\frac{y^2}{x^2} \) ......(1)
Put \( y = vx \), i.e. \( v = \frac{y}{x} \)
\( \implies \frac{dy}{dx} = v+x\frac{dv}{dx} \)
\( \implies \) (1) becomes,
\( v+x\frac{dv}{dx} = 1+v-v^2 \)
\( \implies x\frac{dv}{dx} = 1-v^2 \)
\( \implies \frac{dv}{1-v^2} = \frac{dx}{x} \)
Integrating, we get
\( \int \frac{dv}{1-v^2} = \int \frac{dx}{x} \)
\( \implies \frac{1}{2} \log \left| \frac{1+v}{1-v} \right| = \log|x|+\log c_1 \)
\( \implies \log \left| \frac{1+v}{1-v} \right| = 2\log|x|+2\log c_1 \)
\( \implies \log \left| \frac{1+v}{1-v} \right| = \log(x^2c_1^2) \)
\( \implies \frac{1+v}{1-v} = x^2c_1^2 \)
Let \( c = c_1^2 \)
\( \implies \frac{1+v}{1-v} = cx^2 \)
Substitute \( v = \frac{y}{x} \):
\( \implies \frac{1+\frac{y}{x}}{1-\frac{y}{x}} = cx^2 \)
\( \implies \frac{\frac{x+y}{x}}{\frac{x-y}{x}} = cx^2 \)
\( \implies \frac{x+y}{x-y} = cx^2 \), \( c = c_1^2 \), is the required solution.
In simple words: This homogeneous differential equation is solved by substituting \( y=vx \). We separate the variables, integrate using the standard integral form for \( \frac{1}{a^2-x^2} \), and then substitute \( v=y/x \) back to get the general solution.

🎯 Exam Tip: Remember the integral formula for \( \int \frac{1}{a^2-x^2} dx = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| \). Pay attention to algebraic simplifications when substituting \( v=y/x \) to arrive at the final simplified form.