Question

The solution of $y-x\left(\frac{dy}{dx}\right)=a(y^2+\frac{dy}{dx})$ is

• A. $(x+a)(y+a)=cy$
• B. $(x+a)(y-a)=cy$
• C. $(x+a)(y+\frac{1}{a})=cy$
• D. $(x+a)(1-ay)=cy$

Answer 1

The correct option is D $(x+a)(1-ay)=cy$

$y-x\left(\frac{dy}{dx}\right)=a(y^2+\frac{dy}{dx})$

$\Rightarrow y-x\frac{dy}{dx}={ay}^2+a\frac{dy}{dx}$

$\Rightarrow y-{ay}^2=\frac{dy}{dx}(a+x)$

$\Rightarrow \frac{y-{ay}^2}{a+x}=\frac{dy}{dx}$

$\Rightarrow \frac{dx}{a+x}=\frac{dy}{y-ayz}$

integrating both sides we get,

$\Rightarrow \int \frac{dx}{a+x}=\int \frac{dy}{ay-{ay}^2}$

$\Rightarrow \int \frac{dx}{a+x}=\int \frac{dy}{y(1-ay)}$

$\Rightarrow \int \frac{dx}{a+x}=\int [\frac{1}{y}+\frac{a}{1-ay}]dy$

$\Rightarrow \int \frac{dx}{a+x}=\int \frac{1}{y}dy+\int \frac{ady}{1-ay}$

$\Rightarrow ln(a+x)=lny-ln(1-ay)-in{c}^1$

$\Rightarrow ln(a+x)=ln\left(\frac{y}{(1-ay)c}\right)$

$\Rightarrow a+x=\frac{y}{c^1(1-ay)}$

$\Rightarrow (x+a)(1-ay)=\frac{y}{c^1}$

$(x+a)(1-ay)=cy$ where $c=\frac{1}{c^1}$