# Thread: Finding a particular solution using the Method of Undetermined Coefficients

1. ## Finding a particular solution using the Method of Undetermined Coefficients

Hi all,

I have been trying to solve this problem for hours and have looked at many examples, but I still can't solve it. It is a past exam question, so I don't know if there was an error or if I am just doing it wrong.

The problem says:

"Consider the ordinary differential equation

$y''-2y'+y=te^t+4$

Using the Method of Undetermined Coefficients, find a particular solution $y_P(t)$ of the inhomogeneous problem."

So I have tried to solve it using $y_P=Ate^t+Be^t+C=e^t(At+B)+C$, but I can't get anything because the only result I get is $C=te^t+4$, and no values for $A$ or $B$. And $y_P=te^t+4$ would give $y''-2y'+y=4$ instead of $te^t+4$.

I would really appreciate it if someone could give me a hand. Thanks a lot!

2. ## Re: Finding a particular solution using the Method of Undetermined Coefficients

You are not using the correct form for the particular solution, because the homogeneous solution is

$y_h(t)=c_1e^t+c_2te^t$

Hence, you need to try a particular solution of the form:

$y_p(t)=t^2(At+B)e^t+C$

Do you understand why?

3. ## Re: Finding a particular solution using the Method of Undetermined Coefficients

Thanks a lot! I actually don't understand why though... Where does $t^2$ come from?

4. ## Re: Finding a particular solution using the Method of Undetermined Coefficients

No term in the particular solution can be a solution to the corresponding homogeneous solution, and so that's why the factor $t^2$ is necessary. Have you studied the annihilator method yet?

5. ## Re: Finding a particular solution using the Method of Undetermined Coefficients

No, I haven't... But I've just googled it and found some good files. Thanks a lot!

6. ## Re: Finding a particular solution using the Method of Undetermined Coefficients

If we observe that:

$\frac{d}{dt}\left(te^t+4 \right)-\left(te^t+4 \right)=e^t-4$

$\frac{d}{dt}\left(e^t-4 \right)-\left(e^t-4 \right)=4$

$\frac{d}{dt}\left(4)=0$

we may then state that the differential operator:

$A\equiv D(D-1)^2$

annihilates $te^t+4$

Now, applying this to the given ODE, we have:

$D(D-1)^4[y]=0$

Hence, the general solution must be of the form:

$y(t)=c_1+c_2e^t+c_3te^t+c_4t^2e^t+c_5t^3e^t$

But, we see that the homogeneous solution to the original ODE is:

$y_h(t)=c_2e^t+c_3te^t$

and we know the general solution is the superposition of the homogeneous and particular solutions:

$y(t)=y_h(t)+y_p(t)$

Hence, we must conclude that the particular solution is of the form:

$y_p(t)=c_1+c_4t^2e^t+c_5t^3e^t=t^2\left(c_4+c_5t \right)e^t+c_1$

7. ## Re: Finding a particular solution using the Method of Undetermined Coefficients

Sorry, I was out of town...

We had not been taught about that, but thanks so much! That makes everything a lot clearer now.