Question

The solution of $x^{2}y-x^3\frac{dy}{dx}=y^4\cos x$, given that $x=\pi ,y=\pi$ is

• A. $y^3\left(1+3\sin x\right)=x^3$
• B. $x^3\left(1+3\sin x\right)=y^3$
• C. $y^3\left(1-\sin x\right)=x^3$
• D. $x^3\left(1-\sin x\right)=y^3$

Answer 1

The correct option is A $y^3\left(1+3\sin x\right)=x^3$

$x^{2}y-x^3\frac{dy}{dx}=y^{4}cosx$

$\Rightarrow \frac{x^2}{y^3}-\frac{x^3}{y^4}\frac{dy}{dx}=cosx$

$\Rightarrow \frac{1}{3}(\frac{3x^2.y^3}{y^6}-\frac{x^3.3y^2}{y^6}\frac{dy}{dx})=cosx$

$\Rightarrow \frac{3x^2{y}^3}{y^6}-\frac{x^3.3y^2}{y^6}\frac{dy}{dx}=3cosx$

$\Rightarrow \frac{d}{dx}\left(\frac{x^3}{y^3}\right)=3cosx$

$\Rightarrow \int d\left(\frac{x^3}{y^3}\right)=\int 3cosxdx$

$\Rightarrow \frac{x^3}{y^3}=3sinx+c$

at $x=\pi ,y=x\pi$

$\Rightarrow \frac{{\pi }^3}{{\pi }^3}=3.0+c\Rightarrow c=1$

$\therefore \frac{x^3}{y^3}=3sinx+1$

$\Rightarrow y^3(1+3sinx)=x^3.$