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iRaid
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Mod note: Reinstated problem after poster deleted it.
Just wondering if I did this correctly: ##y''+4y'+4y=e^{x}## and initial conditions ##y(0)=0; y'(0)=1##
So I found the characteristic equation to be ##r^{2}+4r+4=0## so r=-2 and the general solution is then: ##y_{g}=c_{1}e^{-2x}-c_{2}xe^{-2x}## and particular solution: ##y_{p}=Ae^{x}## and obviously the first and second derivatives are going to be the same thing. So plugging the particular solution into the problem: ##Ae^{x}+4Ae^{x}+4Ae^{x}=e^{x}##, so ##A=\frac{1}{9}##.
Now ##y=y_{g}+y_{p}## Which is: ##y=c_{1}e^{-2x}+c_{2}xe^{-2x}+\frac{1}{9}e^{x}##. Finally taking initial conditions y(0)=0 y'(0)=1 I get: $$y=\frac{-1}{9}e^{-2x}+\frac{-7}{9}xe^{-2x}+\frac{1}{9}e^{x}$$
Homework Statement
Just wondering if I did this correctly: ##y''+4y'+4y=e^{x}## and initial conditions ##y(0)=0; y'(0)=1##
Homework Equations
The Attempt at a Solution
So I found the characteristic equation to be ##r^{2}+4r+4=0## so r=-2 and the general solution is then: ##y_{g}=c_{1}e^{-2x}-c_{2}xe^{-2x}## and particular solution: ##y_{p}=Ae^{x}## and obviously the first and second derivatives are going to be the same thing. So plugging the particular solution into the problem: ##Ae^{x}+4Ae^{x}+4Ae^{x}=e^{x}##, so ##A=\frac{1}{9}##.
Now ##y=y_{g}+y_{p}## Which is: ##y=c_{1}e^{-2x}+c_{2}xe^{-2x}+\frac{1}{9}e^{x}##. Finally taking initial conditions y(0)=0 y'(0)=1 I get: $$y=\frac{-1}{9}e^{-2x}+\frac{-7}{9}xe^{-2x}+\frac{1}{9}e^{x}$$
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