Question

Найти частное решение диффереального уравнения​

Answer 1

$y''+y'-6y=e^{2x}\\\\a)\ \ k^2+k-6=0\ \ ,\ \ k_1=-3\ ,\ k_2=2\\\\y_{obshee\ odnor.}=C_1e^{-3x}+C_2e^{2x}\\\\b)\ \ f(x)=e^{2x}\ \ ,\ \ 2=k_2\ \ \to \ \ r=1\\\\\widetilde {y}=Ax^{r}e^{2x}=Axe^{2x}\ \ ,\\\\\widetilde {y}\, '=Ae^{2x}+2Axe^{2a}\\\\\widetilde {y}\, ''=2Ae^{2x}+2Ae^{2x}+4Axe^{2x}\\---------------\\\widetilde {y}\, ''+\widetilde {y}\, '-6\widetilde {y}=5Ae^{2x}=e^{2x}\\\\5A=1\ \ ,\ \ \ A=\dfrac{1}{5}\ \ \ \to \ \ \ \widetilde {y}=\dfrac{1}{5}\, xe^{2x}$

$c)\ \ y_{obshee\ neodn,}=y_{obshee\ odnor,}+\widetilde {y}=C_1e^{-3x}+C_2e^{2x}+\dfrac{1}{5}\, xe^{2x}$