Question

Y"-4y'+4y=0 y1=e^2x find the general solution

Answer 1

Given that
`y_1=e^(2x)` is a known solution to the ODE, we can reduce the order of the ODE to find a second solution of the form


`y_2=y_1v=e^(2x)v`

`\implies {y_2}'=2e^(2x)v+e^(2x)v'`

`\implies {y_2}''=4e^(2x)v+4e^(2x)v'+e^(2x)v''`

Substituting into the ODE, we get


`(e^(2x)v''+4e^(2x)v'+4e^(2x)v)-4(e^(2x)v'+2e^(2x)v)+4e^(2x)v=e^(2x)`

`e^(2x)v''=e^(2x)`

`v''=1`

`\implies v'=C_1`

`\implies v=C_1x+C_2`

`\implies y_2=C_1xe^(2x)+C_2e^(2x)`

We already know about
`e^(2x)` as a solution to the ODE, which means
`y_2=xe^(2x)`.

That covers the characteristic solution. To find the particular solution to the nonhomogeneous ODE, suppose there is a solution of the form


`y_p=ax^2e^(2x)`

`{y_p}'=2ax(x+1)e^(2x)`

`{y_p}''=2a(2x^2+4x+1)e^(2x)`

Substituting into the ODE yields


`{y_p}''-4{y_p}'+4{y_p}=2ae^(2x)=e^(2x)\implies a=\frac12`

so that the general solution is


`y=C_1y_1+C_2y_2+y_p`

`y=C_1e^(2x)+C_2xe^(2x)+\frac12x^2e^(2x)`