Question

8) Use Reduction of order to solve. One solution of homogeneo x2y" +7xy' +5y =x 1 x>0 y1 = X here y1 is a solution of the corresponding homogeneous.

Answer 1

I suspect there's a typo in the question, because
`y_1=x` is *not* a solution to the corresponding homogeneous equation. We have
`{y_1}'=1` and
`{y_1}''=0`, so the ODE reduces to

`0+7x+5x=12x\\eq0`

Let
`y=x^m`, then
`y'=mx^(m-1)` and
`y''=m(m-1)x^(m-2)`, and substituting these into the (homogeneous) ODE gives

`m(m-1)x^m+7mx^m+5x^m=0\implies m(m-1)+7m+5=m^2+6m+5=(m+5)(m+1)=0`

which then admits the characteristic solutions
`y_1=\frac1x` and
`y_2=\frac1{x^5}`.

Now to find a solution to the non-homogeneous ODE. We look for a solution of the form
`y(x)=v(x)y_1(x)` or
`y(x)=v(x)y_2(x)`.

It doesn't matter which one we start with, so let's use the first case. We get derivatives
`y'=x^(-1)v'-x^(-2)v` and
`y''=x^(-1)v''-2x^(-2)v'+2x^(-3)v`. Substituting into the ODE yields

`x^2(x^(-1)v''-2x^(-2)v'+2x^(-3)v)+7x(x^(-1)v'-x^(-2)v)+5x^(-1)v=x`

`xv''+5v'=x`

Substitute
`w=v'`, so that
`w'=v''` and

`xw'+5w=x`

which is linear in
`w`, and we can condense the left side as the derivative of a product after multiplying both sides by
`x^4`:

`x^5w'+5x^4=x^5\implies(x^5w)'=x^5\implies x^5w=\frac{x^6}6+C\implies w=\frac x6+\frac C{x^5}`

Integrate to solve for
`v`:

`v=(x^2)/(12)+(C_1)/(x^4)+C_2`

Then multiply both sides by
`y_1=\frac1x` to solve for
`y`:

`y=\frac x{12}+(C_1)/(x^5)+\frac{C_2}x`

so we found another fundamental solution
`y_3=x` that satisifes this ODE.