Question

The solution of $\frac{dy}{dx}+\frac{y}{x}=x^2{y}^6$ is

• A. $\frac{1}{x^2{y}^5}=\frac{5}{2x^2}+c$
• B. $\frac{1}{x^5{y}^5}=\frac{5}{x^2}+c$
• C. $\frac{1}{x^5{y}^5}=\frac{5}{2x^2}+c$
• D. $\frac{1}{x^2{y}^2}=\frac{5}{2x^2}+c$

Answer 1

Answer: (A)

$\frac{1}{x^2{y}^5}=\frac{5}{2x^2}+c$

This is a first order Bernocilli $ODE$ of type, $y^1+p\left(x\right)y=q\left(x\right)y^n$

$p\left(x\right)=1/x$

$q\left(x\right)=x^2$

$n=6$

Rearrange the ODE as,

$\frac{1}{y^6}\left(\frac{dy}{dx}\right)+\frac{1}{y^{5}x}=x^2$

Perform the substitution

$v=y^{-5}$

$\frac{dv}{dx}=-\frac{5}{y^6}\left(\frac{dy}{dx}\right)$

Therefore, ODE is $\Rightarrow \frac{-1}{5}\frac{dv}{dx}+\frac{v}{x}=x^2$

$\frac{dv}{dx}-\left(\frac{5}{x}\right)v=-5x^2$

Apply the integrating factor,

$e^{(\int -5/x)dx}=e^{-5sinx}=e^{-ln\left(x^5\right)}=1/x^5$

$\therefore \frac{1}{x^5}\frac{dv}{dx}-\frac{1}{x^5}.\left(\frac{5}{x}\right)v=\frac{-1}{x^5}.5x^2$

$\frac{d\left(1/x^{5}v\right)}{dx}=-5/x^3$

Integrating both sides,

$\frac{1}{x^5}v=\frac{5}{2x^2}+C$

$V=\frac{5}{2}{x}^3+C{x}^5$

$\frac{1}{y^5}=\frac{5}{2}{x}^3+C{x}^5=\frac{x^3(5+2C{x}^2)}{2}$

$y^5=\frac{2}{x^3(5+2C{x}^2)}$

$y={\left(\frac{2}{x^3(5+2C{x}^2)}\right)}^{1/5}$

Simplified we get,

$\frac{1}{x^2{y}^5}=\frac{5}{2x^2}+C$