Differential equation Exercise asked by Irfan Maths in the math group

Exercise:

Solve the following Differential equation $\frac{dy}{dx}+\frac{2y}{3}=\frac{x}{\sqrt{y}}$

Solution: we have $\frac{dy}{dx}+\frac{2}{3}y=\frac{x}{\sqrt{y}}\Leftrightarrow \frac{3}{2}\sqrt{y}\frac{dy}{dx}+y\sqrt{y}=\frac{3}{2}x$

Let $z=y\sqrt{y}\Leftrightarrow \frac{dz}{dx}=\frac{3}{2}\sqrt{y}\frac{dy}{dx}$ so $\frac{dz}{dx}+z=\frac{3}{2}x$

Now solving for $\frac{dz}{dx}+z=0\Leftrightarrow dz=-zdx\Leftrightarrow \frac{dz}{z}=-dx\Leftrightarrow \ln z=-x+c\Leftrightarrow z=c{{e}^{-x}}$

Let ${{z}_{p}}=a+bx\Leftrightarrow z{{'}_{p}}=b$ so $b+a+bx=\frac{3}{2}x\Leftrightarrow a+b=0\,\And \,\,b=\frac{3}{2}$

Thus $z\left( x \right)=c{{e}^{-x}}+\frac{3}{2}\left( -1+x \right)$

Hence $z\left( x \right)=\sqrt{{{y}^{3}}}\Leftrightarrow {{z}^{2}}\left( x \right)={{y}^{3}}\Leftrightarrow y=\sqrt[3]{{{z}^{2}}\left( x \right)}$

Thus $y\left( x \right)={{\left( c{{e}^{-x}}+\frac{3}{2}\left( x-1 \right) \right)}^{2/3}}$