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

${y_p}''-4{y_p}'+4{y_p}=2ae^{2x}=e^{2x}\implies a=\dfrac12$

so that the general solution is

$y=C_1y_1+C_2y_2+y_p$
$y=C_1e^{2x}+C_2xe^{2x}+\dfrac12x^2e^{2x}$