Solve the initial value problem x^(2) (d^(2)y)/(dx^(2)) -3x (dy)/(dx) + 20y=0 y(1)=2 y'(1)=-8

Question

Solve the initial value problem
`x^(2) (d^(2)y)/(dx^(2)) -3x (dy)/(dx) + 20y=0`


`y(1)=2`


`y'(1)=-8`

Answer 1

`x^2(\mathrm d^2y)/(\mathrm dx^2)-3x(\mathrm dy)/(\mathrm dx)+20y=0`

This is your standard Euler-Cauchy ODE, which means we can substitute
`y=x^r`, then determine the possible values of
`r` to generate the set of fundamental solutions.


`y=x^r`

`y'=rx^(r-1)`

`y''=r(r-1)x^(r-2)`


`x^2r(r-1)x^(r-2)-3xrx^(r-1)+20x^r=0`

`r(r-1)x^r-3rx^r+20x^r=0`

`r(r-1)-3r+20=0`

`r^2-4r+20=0\implies r=2\pm4i`

For
`r=2+4i`, we get the solution


`x^(2+4i)=x^2x^(4i)=x^2e^(4i\ln x)=x^2(\cos(4\ln x)+i\sin(4\ln x))`

We'll get something almost identical with
`r=2-4i` - namely, two fundamental solutions
`x^2\cos(4\ln x)` and
`x^2\sin(4\ln x)`, and so the general solution will be


`y_c=C_1x^2\cos(4\ln x)+C_2x^2\sin(4\ln x)`

Given
`y(1)=2`, we have


`2=C_1`

and from
`y'(1)=-8` we get


`{y_c}'=2x^2\cos(4\ln x)+C_2x^2\sin(4\ln x)`

`\implies-8=4+4C_2\implies C_2=-3`

so that the particular solution is


`y_c=2x^2\cos(4\ln x)-3x^2\sin(4\ln x)`