Question

Find a second independent solution y1=x xy"-xy'+y=0

Answer 1

We can use reduction of order. Given that
`y_1(x)=x` is a known solution, we look for a solution of the form
`y_2(x)=v(x)y_1(x)`. It has derivatives
`{y_2}'=v'y_1+v{y_1}'` and
`{y_2}''=v''y_1+2v'{y_1}'+v{y_1}''`. Substituting these into the ODE gives

`x(xv''+2v')-x(xv'+v)+xv=0`

`x^2v''+(2x-x^2)v'=0`

Let
`w(x)=v'(x)` so that
`w'(x)=v''(x)` and we get an ODE linear in
`w`:

`x^2w'+(2x-x^2)w=0`

Divide both sides by
`e^x`:

`x^2e^(-x)w'+(2x-x^2)e^(-x)w=0/tex]</p><p>Since [tex](x^2e^(-x))=(2x-x^2)e^(-x)`, we can condense the left side as the derivative of a product:

`(x^2e^(-x)w)'=0`

Integrate both sides and solve for
`w(x)`:

`x^2e^(-x)w=C\implies w=(Ce^x)/(x^2)`

Integrate both sides again to solve for
`v(x)`. Unfortunately, there is no closed form for the integral of the right side, but we can leave the result in the form of a definite integral:

`v=\displaystyle C_2+C_1\int_(x_0)^x(e^t)/(t^2)\,\mathrm dt`

where
`x_0` is any point on an interval over which a solution to the ODE exists.

Finally, multiply by
`y_1(x)` to solve for
`y_2(x)`:

`y_2=\displaystyle C_2x+C_1x\int_(x_0)^x(e^t)/(t^2)\,\mathrm dt`

`y_1(x)` already accounts for the
`C_2x` term above, so the second independent solution is

`y_2=x\displaystyle\int_(x_0)^x(e^t)/(t^2)\,\mathrm dt`