# Solve. dy/dx+3/x y = 27y^(1/3) 1n(x), x > 0

Solve. $\frac{dy}{dx}+\frac{3}{x}y=27{y}^{\frac{1}{3}}1n\left(x\right),x>0$
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Talisha
Here the given differential equation is
${y}^{{}^{\prime }}+\frac{3y}{x}=27{y}^{\frac{1}{3}}$
Let $v={y}^{\frac{2}{3}}then\frac{dv}{dx}=\frac{2}{3}{y}^{\frac{1}{3}}\frac{dy}{dx}$.With respect to this transformation, the above DE reduces to $\frac{3}{2}{y}^{\frac{1}{3}}\frac{dv}{dx}+\frac{3y}{x}=27{y}^{\frac{1}{3}}$
$⇒\frac{3}{2}\frac{dv}{dx}+\frac{3{y}^{\frac{2}{3}}}{x}=27$
$⇒\frac{1}{2}\frac{dv}{dx}+\frac{{y}^{\frac{2}{3}}}{x}=9$
$⇒\frac{dv}{dx}+\frac{2{y}^{\frac{2}{3}}}{x}=18$
$⇒\frac{dv}{dx}+2\frac{v}{x}=18$
$⇒{x}^{2}\frac{dv}{dx}+2xv=18{x}^{2}$
$⇒\frac{d}{dx}\left({x}^{2}v\right)=18{x}^{2}$
$⇒{x}^{2}v=\int 18{x}^{2}dx+c$
$⇒{x}^{2}v=6{x}^{3}+c$
$⇒v=6x+c{x}^{-2}$
$⇒{y}^{\frac{2}{3}}=6x+c{x}^{-2}$
$⇒y={\left(6x+c{x}^{-2}\right)}^{\frac{3}{2}}$
Where c is arbitrary constant.