Question

Solution of the differential equation $x^{2}y-x^3\frac{dy}{dx}=y^4\cos x$ is
[Note: where C is constant of integration.]

• A. $x^2{y}^{-3}=2\sin x+C$
• B. $x^2{y}^{-3}=3\cos x+C$
• C. $x^3{y}^{-3}=3\sin x+C$
• D. $x^2{y}^3=3\sin x+C{x}^{2}y$

Answer 1

Answer: (C)

$\Theta x^{2}y-x^3\frac{dy}{dx}=y^4\cos x$
$\Rightarrow \frac{-1}{y^4}\frac{dy}{dx}+\frac{1}{x}\cdot \frac{1}{y^3}=\frac{\cos x}{x^3}$
$\Rightarrow \frac{-3}{y^4}\frac{dy}{dx}+\frac{3}{x}\cdot \frac{1}{y^3}=\frac{3\cos x}{x^3}$
Let $\frac{1}{y^3}=t\Rightarrow \frac{-3}{y^4}\frac{dy}{dx}=\frac{dt}{dx}$
$\therefore \frac{dt}{dx}+\frac{3}{x}\cdot t=\frac{3\cos x}{x^3}$
$\text{ I.F. }=e^{\int \frac{3}{x}dx}=e^{3\ln x}=x^3$
$\therefore$ solution will be
$t\cdot x^3=\int \frac{3\cos x}{x^3}\cdot x^{3}dx$
$\Rightarrow x^3{y}^{-3}=3\sin x+C.$