Solving 2nd Order Differential Equation with Initial Conditions

[iRaid]

Mod note: Reinstated problem after poster deleted it. [/color]

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: $$y=\frac{-1}{9}e^{-2x}+\frac{-7}{9}xe^{-2x}+\frac{1}{9}e^{x}$$

 

[Dick]

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: $$y=\frac{-1}{9}e^{-2x}+\frac{-7}{9}xe^{-2x}+\frac{1}{9}e^{x}$$

You can check these things yourself. Substitute your solution into the DE and conditions you are given. That's what the checkers you enlisted will do. I agree that y satisfies the DE and that y(0)=0. I don't think y'(0)=1.

 

[iRaid]

You can check these things yourself. Substitute your solution into the DE and conditions you are given. That's what the checkers you enlisted will do. I agree that y satisfies that DE and the y(0)=0. I don't think y'(0)=1.

I'm not so much concerned with the answer, I want to know if I'm doing the work correctly and whether or not there's a simpler way to solve it.

 

[Dick]

I'm not so much concerned with the answer, I want to know if I'm doing the work correctly and whether or not there's a simpler way to solve it.

You are doing it correctly and I don't think there's simpler way. You just goofed up a little in solving for c1 and c2 in applying the boundary conditions.

 

[iRaid]

You are doing it correctly and I don't think there's simpler way. You just goofed up a little in solving for c1 and c2 in applying the boundary conditions.

OK that's all I was wondering thanks.

 

[Dick]

done thanks

Now why did you delete the OP? Part of the use of Physics Forums is to provide a resource for people to look up past solutions and get some hints for their own problem. Deleting parts of threads makes them unreadable. That's, in part, why I quoted you. That makes the deletion doubly pointless.

 

[Mark44]

iRaid, For the reason that Dick gave, please don't delete your post just because you got an answer.