Answer:
[tex]y=(x^5/8+Cx^{-3})^{3/2}[/tex]
Step-by-step explanation:
xy’+3y=x^5y^(1/3)
rewrite in:
[tex]y^{-1/3}\frac{dy}{dx} +3x^{-1}y^{2/3}=x^4[/tex]
substitute: [tex]v=y^{2/3}\\[/tex]
[tex]\frac{dv}{dx}=\frac{d}{dx}y^{2/3}=y^{-1/3}\frac{dy}{dx}[/tex]
The equation can be rewritten in:
[tex]\frac{dv}{dx}+3vx^{-1}=x^4[/tex]
This is a first-order linear differential equation:
Can be solved by using integrating factor or variation of parameter:
integrating factor:
[tex]u(x)=exp([/tex]∫[tex]3/x \ dx) = x^3[/tex]
[tex]v=u^{-1}[/tex]∫[tex]u(x) x^4dx[/tex]
[tex]v=x^{-3}([/tex]∫[tex]x^4x^3dx)[/tex]
[tex]y^{2/3}=v=x^{-3}(x^8/8+C)[/tex]
[tex]y=(x^5/8+Cx^{-3})^{3/2}[/tex]
