Non-homogeneous differential equation, undetermined coefficients

[ToastIQ]

Hi!
I've been stuck on this problem for 2 days. I'm new here and I've just spent another hour on trying to figure out Latex, but it always ends up in a mess so I'll try without. I hope that's okay.

The equation is

y''+6y'+9y=4e^-x

with the value boundaries (I think that's what it's called in english?) y(0)=2 and y'(0)=-2.

I got the complementary solution to y_c=C₁e^-3x+C₂e^-3x.

I'm assuming my mistake is somewhere after this. I set up the particular solution:

y_p=Ae^-x and get
y'_p=-Ae^-x and y''_p=Ae^-x.
I get A=1. Combining complementary and particular solution:

y=C₁e^-3x+C₂e^-3x+e^-x

I can't seem to get the right coefficients. The result should be

y=(1+2x)e^-3x+e^-x.

I've tried with and without the 4 in 4e^-x, I've tried multiplying by x to get a higher degree and several other alternatives. I've been searching the internet for the most of the past 48 hours. I read somewhere that e^-x is a "special case" but no example for this case was given. My textbook doesn't touch upon this at all. It would be very appreciated if someone could help me out.

Thanks in advance!

 

[MarkFL]

Hello and welcome to MHB! (Wave)

The first thing I notice when reviewing your work, is that you have a repeated root in your characteristic equation, and so the homogeneous solution should be:

\(\displaystyle y_h(x)=c_1e^{-3x}+c_2xe^{-3x}\)

See if that gets you to the correct solution. :)

 

[ToastIQ]

Hello and welcome to MHB! (Wave)

The first thing I notice when reviewing your work, is that you have a repeated root in your characteristic equation, and so the homogeneous solution should be:

\(\displaystyle y_h(x)=c_1e^{-3x}+c_2xe^{-3x}\)

See if that gets you to the correct solution. :)

Thank you for your reply.

I actually remember trying this alternative too. This gave me the right coefficient for C₁ but I still got it wrong for C₂. I tried again and saw I had the wrong derivative for C₂xe^-3x. That solved the problem.
Thanks a lot! :)

 

[MarkFL]

I've moved this thread here to our "Differential Equations" forum as it is a better fit, and for the benefit of guests who may visit I'm going to provide a full solution.

We are given to solve the IVP:

\(\displaystyle y''+6y'+9y=4e^{-x}\) where \(\displaystyle y(0)=2,\,y'(0)=-2\)

As discussed above, you homogeneous solution is:

\(\displaystyle y_h(x)=c_1e^{-3x}+c_2xe^{-3x}\)

And our particular solution will take the form:

\(\displaystyle y_p(x)=Ae^{-x}\)

Hence:

\(\displaystyle y_p'(x)=-Ae^{-x}\)

\(\displaystyle y_p''(x)=Ae^{-x}\)

Substituting into the given ODE, we find:

\(\displaystyle Ae^{-x}-6Ae^{-x}+9Ae^{-x}=4e^{-x}\)

Multiply through by $e^x$, since $e^x\ne0$:

\(\displaystyle A-6A+9A=4\)

\(\displaystyle 4A=4\implies A=1\)

And so our particular solution is:

\(\displaystyle y_p(x)=e^{-x}\)

Now, by the principle of superposition, we obtain:

\(\displaystyle y(x)=y_h(x)+y_p(x)=c_1e^{-3x}+c_2xe^{-3x}+e^{-x}\)

Hence:

\(\displaystyle y'(x)=-3c_1e^{-3x}-3c_2xe^{-3x}+c_2e^{-3x}-e^{-x}\)

Now, using general initial values, we may write:

\(\displaystyle y(0)=c_1+1=y_0\implies c_1=y_0-1\)

\(\displaystyle y'(0)=-3c_1+c_2-1=-3\left(y_0-1\right)+c_2-1=2-3y_0+c_2=y'_0\implies c_2=3y_0+y'_0-2\)

And so the general solution is:

\(\displaystyle y(x)=\left(y_0-1\right)e^{-3x}+\left(3y_0+y'_0-2\right)xe^{-3x}+e^{-x}\)

Plugging in the given initial values, we obtain the solution to the IVP:

\(\displaystyle y(x)=\left(2-1\right)e^{-3x}+\left(3(2)+(-2)-2\right)xe^{-3x}+e^{-x}=e^{-3x}+2xe^{-3x}+e^{-x}\)