Question

Solve the following differential equation :$xdy-y-x{y}^3(1+lnx)dx=0$

• A. $\frac{x^2}{y^2}=-\frac{2}{3^3}(\frac{2}{3}+lnx)+c$
• B. $\frac{x^2}{y^2}=\frac{2}{3^3}(\frac{2}{3}+lnx)+c$
• C. $\frac{x^2}{y}=\frac{2}{3^3}(\frac{2}{3}-lnx)+c$
• D. $\frac{x^2}{y}=\frac{2^2}{3^3}(\frac{2}{3}-lnx)+c$

Answer 1

Answer: (A)

$\frac{x^2}{y^2}=-\frac{2}{3^3}(\frac{2}{3}+lnx)+c$

$-y+x\frac{dy}{dx}-x{y}^3(\ln x+1)=0\Rightarrow x\frac{dy}{dx}-y=-\left(x(-\ln x-1)y^3\right)\Rightarrow -\frac{2\frac{dy}{dx}}{y^3}+\frac{2}{x{y}^2}=-2(\ln x+1)$
Let $v=\frac{1}{y^2}\Rightarrow \frac{dv}{dx}=-\frac{2\frac{dy}{dx}}{y^3}$
$\frac{dv}{dx}+\frac{2v}{x}=-2(\ln x+1)$
Let $\mu =e^{\int \frac{2}{x}dx}=x^2$
$x^2\frac{dv}{dx}+2xv=-2x^2(\ln x+1)$
Substitute $2x=\frac{d}{dx}\left(x^2\right)$
$x^2\frac{dv}{dx}+\frac{d}{dx}\left(x^2\right)v=-2x^2(\ln x+1)$
Using $g\frac{df}{dx}+f\frac{dg}{dx}=\frac{d}{dx}\left(fg\right)$
$\frac{d}{dx}\left(x^{2}v\right)=-2x^2(\ln x+1)\Rightarrow \int \frac{d}{dx}\left(x^{2}v\right)dx=\int -2x^2(\ln x+1)dx\Rightarrow x^{2}v=-\frac{4x^3}{9}-\frac{2}{3}{x}^3\ln x+c\Rightarrow v=-\frac{4x}{9}-\frac{2}{3}x\ln x+\frac{c}{x^2}$
Hence $\Rightarrow \frac{x^2}{y^2}=-\frac{2}{3^3}(\frac{2}{3}+lnx)+c$