Question

Bernoulli equation xy’+3y=x^5y^(1/3) Please solve this step by step

Answer 1

Answer:

$y=(x^5/8+Cx^{-3})^{3/2}$

Step-by-step explanation:

xy’+3y=x^5y^(1/3)

rewrite in:

$y^{-1/3}\frac{dy}{dx} +3x^{-1}y^{2/3}=x^4$

substitute: $v=y^{2/3}$

$\frac{dv}{dx}=\frac{d}{dx}y^{2/3}=y^{-1/3}\frac{dy}{dx}$

The equation can be rewritten in:

$\frac{dv}{dx}+3vx^{-1}=x^4$

This is a first-order linear differential equation:

Can be solved by using integrating factor or variation of parameter:

integrating factor:

$u(x)=exp($∫$3/x \ dx) = x^3$

$v=u^{-1}$∫$u(x) x^4dx$

$v=x^{-3}($∫$x^4x^3dx)$

$y^{2/3}=v=x^{-3}(x^8/8+C)$

$y=(x^5/8+Cx^{-3})^{3/2}$