Respuesta :

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Given that [tex]y_1=e^{2x}[/tex] is a known solution to the ODE, we can reduce the order of the ODE to find a second solution of the form

[tex]y_2=y_1v=e^{2x}v[/tex]
[tex]\implies {y_2}'=2e^{2x}v+e^{2x}v'[/tex]
[tex]\implies {y_2}''=4e^{2x}v+4e^{2x}v'+e^{2x}v''[/tex]

Substituting into the ODE, we get

[tex](e^{2x}v''+4e^{2x}v'+4e^{2x}v)-4(e^{2x}v'+2e^{2x}v)+4e^{2x}v=e^{2x}[/tex]
[tex]e^{2x}v''=e^{2x}[/tex]
[tex]v''=1[/tex]
[tex]\implies v'=C_1[/tex]
[tex]\implies v=C_1x+C_2[/tex]
[tex]\implies y_2=C_1xe^{2x}+C_2e^{2x}[/tex]

We already know about [tex]e^{2x}[/tex] as a solution to the ODE, which means [tex]y_2=xe^{2x}[/tex].

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

[tex]y_p=ax^2e^{2x}[/tex]
[tex]{y_p}'=2ax(x+1)e^{2x}[/tex]
[tex]{y_p}''=2a(2x^2+4x+1)e^{2x}[/tex]

Substituting into the ODE yields

[tex]{y_p}''-4{y_p}'+4{y_p}=2ae^{2x}=e^{2x}\implies a=\dfrac12[/tex]

so that the general solution is

[tex]y=C_1y_1+C_2y_2+y_p[/tex]
[tex]y=C_1e^{2x}+C_2xe^{2x}+\dfrac12x^2e^{2x}[/tex]

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