Помогите решить y"+4y=3e^2x

Решите задачу:

$y''+4y=3e^{2x}\\\\1)\ \ k^2+4=0\ \ ,\ \ k^2=-4\ \ ,\ \ k_{1,2}=\pm 2i\\\\y_{obsh.odn.}=C_1\, cos2x+C_2\, sin2x\\\\2)\ \ f(x)=3e^{2x}\ \ ,\ \ \alpha =2\ne k_{1,2}\ \ \to \ \ r=0\\\\y_{chastn.neodn.}=\widetilde {y}=Ae^{2x}\cdot x^{r}=Ae^{2x}\\\\\widetilde {y}\, '=2Ae^{2x}\ \ \ ,\ \ \ \widetilde {y}\, ''=4Ae^{2x}\\\\\widetilde {y}\, ''+4\widetilde {y}=4Ae^{2x}+4\cdot Ae^{2x}=8Ae^{2x}=3e^{2x}\ \ \ \to \ \ \ 8A=3\ \ ,\ \ A=\dfrac{3}{8}\\\\\widetilde {y}=\dfrac{3}{8}\, e^{2x}$

$y_{obsh.neodn.}=C_1\, cos2x+C_2\, sin2x+\dfrac{3}{8}\, e^{2x}$