Question

Find general solution of $y-x\frac{dy}{dx}=b(1+x^2\frac{dy}{dx})$ is:

• A. $b+kx=y(1+bx)$
• B. $b+ky=x(1+bx)$
• C. $b+ky=x(1+by)$
• D. $b+kx=x(1+by)$

Answer 1

The correct option is A $b+kx=y(1+bx)$

$y-x\frac{dy}{dx}=b(1+x^2\frac{dy}{dx})$
$\Rightarrow y-b=(x+b{x}^2)\frac{dy}{dx}\Rightarrow \frac{dx}{x(1+bx)}=\frac{dy}{y-b}$
$\Rightarrow \int \left(\frac{1}{x}-\frac{b}{1+bx}\right)dx=\log (y-b)+\log c$
$\Rightarrow \log x-\log (1+bx)=\log (y-b)+\log c$
$\Rightarrow \frac{x}{c}=(1+bx)(y-b)$
$\Rightarrow b+\left(\frac{1}{c}+b^2\right)x=y(1+bx)$
$\Rightarrow b+kx=y(1+bx)$