Question

General solution of $y\frac{dy}{dx}+b{y}^2=a\cos x,0<x<1$ is:

(here $c$ is an arbitrary constant)

• A. $y^2=2a(2b\sin x+\cos x)+c{e}^{-2bx}$
• B. $(4b^2+1)y^2=2a(\sin x+2b\cos x)+c{e}^{-2bx}$
• C. $(4b^2+1)y^2=2a(\sin x+2b\cos x)+c{e}^{2bx}$
• D. $y^2=2a(2b\sin x+\cos x)+c{e}^{2bx}$
  

Answer 1

The correct option is B $(4b^2+1)y^2=2a(\sin x+2b\cos x)+c{e}^{-2bx}$

Let us take $y^2=t$ , which means $dt=2ydy$

Given equation will transform into $\frac{dt}{dx}+2bt=2a\cos x$

The integration factor is $e^{2bx}$

The general solution is $t{e}^{2bx}=\int 2a\cos x{e}^{2bx}dx+c$

$\Rightarrow t{e}^{2bx}=\frac{4ab\cos x+2a\sin x}{4b^2+1}{e}^{2bx}+c$

$\Rightarrow (4b^2+1)y^2=2a(\sin x+2b\cos x)+C{e}^{-2bx}$

Therefore the correct option is $B$