Question

The general solution of the differential equation \( \frac{dy}{dx} = \frac{x+y}{x-y} \) is

• A. \( y-x=cx^2 \)
• B. \( \tan^{-1}\left(\frac{y}{x}\right) = \log\left(cx\sqrt{x^2+y^2}\right) \)
• C. \( x+y=cx^2 \)
• D. \( \tan^{-1}\left(\frac{y}{x}\right) = \log\left(c\sqrt{x^2+y^2}\right) \)

Hint:

For a homogeneous differential equation \( \frac{dy}{dx} = F(\frac{y}{x}) \): 1. Substitute \( y=vx \), so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). 2. The equation becomes separable in \(v\) and \(x\): \( x\frac{dv}{dx} = F(v)-v \). 3. Integrate \( \int \frac{dv}{F(v)-v} = \int \frac{dx}{x} \). 4. Substitute back \( v=y/x \). Remember \( \int \frac{1}{1+u^2}du = \tan^{-1}u \) and \( \int \frac{u}{1+u^2}du = \frac{1}{2}\log(1+u^2) \). Properties of logarithm: \( \frac{1}{2}\log A = \log\sqrt{A} \), \( \log A + \log B = \log(AB) \).

Answer 1

The Correct Option is D

The given differential equation is \( \frac{dy}{dx} = \frac{x+y}{x-y} \).
This is a homogeneous differential equation.
Let \( y = vx \).
Then \( \frac{dy}{dx} = v + x\frac{dv}{dx} \).
Substitute into the equation: \[ v + x\frac{dv}{dx} = \frac{x+vx}{x-vx} = \frac{x(1+v)}{x(1-v)} = \frac{1+v}{1-v} \] \[ x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v - v(1-v)}{1-v} = \frac{1+v-v+v^2}{1-v} = \frac{1+v^2}{1-v} \] Separate variables: \[ \frac{1-v}{1+v^2} dv = \frac{1}{x} dx \] Integrate both sides: \[ \int \frac{1-v}{1+v^2} dv = \int \frac{1}{x} dx \] \[ \int \left(\frac{1}{1+v^2} - \frac{v}{1+v^2}\right) dv = \log|x| + C_1 \] \[ \tan^{-1}v - \frac{1}{2}\int \frac{2v}{1+v^2} dv = \log|x| + C_1 \] \[ \tan^{-1}v - \frac{1}{2}\log(1+v^2) = \log|x| + C_1 \] Substitute back \( v = y/x \): \[ \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\log\left(1+\frac{y^2}{x^2}\right) = \log|x| + C_1 \] \[ \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\log\left(\frac{x^2+y^2}{x^2}\right) = \log|x| + C_1 \] \[ \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}(\log(x^2+y^2) - \log(x^2)) = \log|x| + C_1 \] \[ \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\log(x^2+y^2) + \frac{1}{2}\log(x^2) = \log|x| + C_1 \] Since \( \log(x^2) = 2\log|x| \): \[ \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\log(x^2+y^2) + \log|x| = \log|x| + C_1 \] \[ \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\log(x^2+y^2) = C_1 \] \[ \tan^{-1}\left(\frac{y}{x}\right) = \frac{1}{2}\log(x^2+y^2) + C_1 \] \[ \tan^{-1}\left(\frac{y}{x}\right) = \log\sqrt{x^2+y^2} + C_1 \] Let \( C_1 = \log c \).
(Or \(C_1 = \log c'\) if c is inside log) \[ \tan^{-1}\left(\frac{y}{x}\right) = \log\sqrt{x^2+y^2} + \log c = \log(c\sqrt{x^2+y^2}) \] This matches option (4).