Question

Define Bernoulli's equation hence solve( dy/DX)+y/x=x²y^6?

Answer 1

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

Explanation:

This is a first order Bernouilli ODE of the form

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

`p(x)=1/x`

`q(x)=x^2`

`n=6`

The general solution is obtained by the substitution

`v=y^(1-n)`

and solving the equation

`1/(1-n)v'+p(x)v=q(x)`

Divide both sides by `x^6`

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

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

Let `v=y^-5`

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

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

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

The integrating factor is

`IF=e^(int-5/xdx)=e^(-5lnx)=1/x^5`

Therefore,

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

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

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

`v/x^5=5/2*1/x^2+C`

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

Substitute back the value of `y`

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

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

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