Question

The general solution of the differential equation $x^{2}dy-2xydx=x^4\cos xdx$ is

• A. $y=x^2\sin x+c{x}^2$
• B. $y=x^2\sin x+c$
• C. $y=\sin x+c{x}^2$
• D. $y=\cos x+c{x}^2$

Answer 1

Answer: (A)

$x^{2}dy-2xydx=x^4\cos xdx$
$\left(\frac{dy}{dx}=\frac{x^{4}cosx+2xy}{x^2}\right)$
$\Rightarrow dy/dx-2y/x=x^2\cos x$
I.F. $=e^{\int -2/xdx}=e^{-2\log x}=1/x^2$
Therefore, the general solution is
$\left(y(\frac{1}{x^2})=\int \frac{1}{x^2}(x^{2}cosx)dx=sinx+c\right)$
$\therefore y=x^2(\sin x+c)$
$=x^2\sin x+c{x}^2$