Respuesta :
Answer:
A) Particular solution:
[tex]2x+\frac{1}{2}e^{-x}-\frac{8}{5}[/tex]
B) Homogeneous solution:
[tex]y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))[/tex]
C) The most general solution is
[tex]y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}[/tex]
Step-by-step explanation:
Given non homogeneous ODE is
[tex]y''+4y'+5y=10x+e^{-x}---(1)[/tex]
To find homogeneous solution:
[tex]D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)[/tex]
To find particular solution:
[tex]y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\[/tex]
Substituting [tex]y_{p},y'_{p},y''_{p}[/tex] in (1)
[tex]y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\[/tex]
Equating the coefficients
[tex]5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\[/tex]
The general solution is
[tex]y=y_{h}+y_{p}[/tex]
from (2) ad (3)
[tex]y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}[/tex]
