Question

Solve the following differential equations:

$\frac{dy}{dx}+y=y^2(\cos x-\sin x)$

• A. $y=\frac{1}{-\sin x+c{e}^x}$
• B. $y=\frac{1}{\sin x+c{e}^x}$
• C. $y=\frac{1}{\sin x+c{e}^{-x}}$
• D. $y=\frac{1}{\cos x+c{e}^{-x}}$

Answer 1

The correct option is A $y=\frac{1}{-\sin x+c{e}^x}$

Given differential equation $\frac{dy}{dx}+y=y^2(\cos x-\sin x)$

$y^{-2}\frac{dy}{dx}+y^{-1}=(\cos x-\sin x)$

Let $v=y^{-1}$

$\frac{dv}{dx}=-y^{-2}\frac{dy}{dx}$

Now,

$-\frac{dv}{dx}+v=(\cos x-\sin x)$

$\frac{dv}{dx}-v=(\sin x-\cos x)$

I.F (Integrating Factor) $=e^{\int -dx}=e^{-x}$

$I.F\times v=\int I.F.\times (\sin x-\cos x)dx$

$e^{-x}v=\int e^{-x}(\sin x-\cos x)dx$

$e^{-x}v=\int e^{-x}\sin xdx-\int e^{-x}\cos xdx$

$e^{-x}v=\int e^{-x}\sin xdx-\int e^{-x}\sin xdx-e^{-x}\sin x+c$

$e^{-x}v=-e^{-x}\sin x+c$

$v=-\sin x+c{e}^x$

$\frac{1}{y}=-\sin x+c{e}^x$

$y=\frac{1}{-\sin x+c{e}^x}$