Question

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

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

Answer 1

Answer: (A)

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