Question

Solution of the differential equation $x^2 y-x^3 \frac{d y}{d x}=y^4 \cos x$ 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)

$\Theta x^2 y-x^3 \frac{d y}{d x}=y^4 \cos x$
$\Rightarrow \frac{-1}{y^4} \frac{d y}{d x}+\frac{1}{x} \cdot \frac{1}{y^3}=\frac{\cos x}{x^3} $
$\Rightarrow \frac{-3}{y^4} \frac{d y}{d x}+\frac{3}{x} \cdot \frac{1}{y^3}=\frac{3 \cos x}{x^3}$
Let $\frac{1}{y^3}=t \Rightarrow \frac{-3}{y^4} \frac{d y}{d x}=\frac{d t}{d x}$
$\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 d x $
$\Rightarrow x^3 y^{-3} =3 \sin x+C . $