Question

The solution of differential equation $x \frac{d y}{d x}+y=x^2 y^4$ is

• A. $\frac{1}{y^3}=3 x^2+c x^3$
• B. $3 x ^2+ y ^3= c$
• C. $x^2=y^3+c$
• D. $y^3=x+c$

Answer 1

Answer: (A)

$x \frac{d y}{d x}+y=x^2 y^4$
$\Rightarrow \frac{1}{y^4} \frac{d y}{d x}+\frac{y}{x} \cdot \frac{1}{y^4}=\frac{x^2}{x}$
$\Rightarrow \frac{1}{y^4} \frac{d y}{d x}+\frac{1}{x y^3}=x \,\,\,\,\frac{1}{y^3}=t \Rightarrow \frac{-3}{y^4} \frac{d y}{d x}=\frac{d t}{d x}$
$\Rightarrow-\frac{1}{3} \frac{d t}{d x}+\frac{t}{x}=x$
$\Rightarrow \frac{d t}{d x}-\frac{3 t}{x}+3 x=0$
I.F. $=e^{-\int \frac{3}{x} d x}=e^{-\ell n x^3}=\frac{1}{x^3}$
$\Rightarrow-\frac{1}{3 x^3} \frac{d t}{d x}+\frac{t}{x^4}=\frac{1}{x^2}$
$\Rightarrow d\left(\frac{-t}{3 x^3}\right)=\frac{1}{x^2} d x$
$\Rightarrow-\frac{t}{3 x^3}=-\frac{1}{x}+c$
$\Rightarrow t=3 x^2-3 c x^3$ $\left(\because t=\frac{1}{y^3}\right)$
$\Rightarrow \frac{1}{y^3}=3 x^2+k x^3$