Question

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

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

The given differential equation is
$y\frac{dy}{dx}+b{y}^2=a\cos x...\left(1\right)$
Let $y^2=t$ then,
$2y\frac{dy}{dx}=\frac{dt}{dx}$
Substitute it in equation (1)
$\frac{1}{2}\frac{dt}{dx}+bt=a\cos x$
$\Rightarrow \frac{dt}{dx}+2bt=2a\cos x$
The integrating factor (I.F.)
$=e^{\int 2.b.dx}=e^{2bx}$
The solution of differential equation is calculated as
$t.e^{2bx}=\int 2a\cos x.e^{2bx}dx$
$\Rightarrow t.e^{2bx}=\frac{2a}{4b^2+1}(\sin x+2b\cos x)e^{2bx}+c$
Now, putting $y^2=t$
$\Rightarrow y^2{e}^{2bx}=\frac{2a}{4b^2+1}(\sin x+2b\cos x)e^{2bx}+c$
$\Rightarrow (4b^2+1)y^2=2a(\sin x+2b\cos x)+c{e}^{-2bx}$