Question

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

• A. $y^3{x}^{-3}=3cosx-c$
• B. $x^3{y}^{-3}=3cosx-c$
• C. $x^3{y}^{-3}=3sinx-c$
• D. $y^3{x}^{-3}=3sinx-c$

Answer 1

The correct option is C $x^3{y}^{-3}=3sinx-c$

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

Let $v=\frac{1}{y^3}\Rightarrow \frac{dv}{dx}=-\frac{3\frac{dy}{dx}}{y^4}$

$\therefore \frac{dv}{dx}+\frac{3v}{x}=\frac{3\cos x}{x^3}$

Let $\mu \left(x\right)=e^{\int \frac{3}{x}dx}=x^3$

Multiply both sides by $\mu \left(x\right)$

$\frac{x^{3}dv}{dx}+\left({3x}^2\right)v=3\cos x\Rightarrow \frac{x^{3}dv}{dx}+\frac{d}{dx}\left(x^3\right)v=3\cos x$

Applying $g\frac{df}{dx}+f\frac{dg}{dx}=\frac{d}{dx}\left(fg\right)$

$\frac{d}{dx}\left(x^{3}v\right)=3\cos x\Rightarrow \int \frac{d}{dx}\left(x^{3}v\right)dx=\int 3\cos xdx\Rightarrow x^{3}v=3\sin x+c\Rightarrow x^3{y}^{-3}=3\sin x+c$