Question

`dy/dx+y/x=x^2y^6`?

Answer 1

The solution is `y=(2/(x^3(5+2Cx^2)))^(1/5)`

Explanation:

This a first order Bernouilli ODE of the type

`y'+p(x)y=q(x)y^n`

`p(x)=1/x`

`q(x)=x^2`

`n=6`

Rearrange the ODE as

`1/y^6(dy/dx)+1/(y^5x)=x^2`

Perform the substitution

`v=y^-5`

`(dv)/dx=-5/y^6(dy/dx)`

Therefore, the ODE is

`-1/5(dv)/dx+v/x=x^2`

`(dv)/dx-(5/x)v=-5x^2`

Apply the integrating factor

`e^((int-5/x)dx)=e^(-5lnx)=e^(-ln(x^5))=1/x^5`

Therefore,

`1/x^5(dv)/dx-1/x^5*(5/x)v=-1/x^5*5x^2`

`1/x^5(dv)/dx-5/x^6v=-5/x^3`

`(d(1/x^5v))/dx=-5/x^3`

Integrating both sides

`1/x^5v=5/(2x^2)+C`

`v=5/2x^3+Cx^5`

`1/y^5=5/2x^3+Cx^5=(x^3(5+2Cx^2))/(2)`

`y^5=2/(x^3(5+2Cx^2))`

`y=(2/(x^3(5+2Cx^2)))^(1/5)`