Use Variation of Parameters to find a particular solution

[shamieh]

Can someone verify that my answer is correct ? Thanks in advance.

Use Variation of Parameters to find a particular solution to $y'' - y = e^t$

Solution:
$y_p = \frac{1}{2}te^t - \frac{1}{4} e^t$

 

[MarkFL]

One of the terms of your particular solution is a solution to the corresponding homogeneous solution.

(a) A fundamental solution set \(\displaystyle \{y_1(t),y_2(t)\}\) for the corresponding homogeneous equation are:

\(\displaystyle y_1(t)=e^{t}\)

\(\displaystyle y_2(t)=e^{-t}\)

and so we take:

\(\displaystyle y_p(t)=v_1(t)y_1(t)+v_2(t)y_2(t)=v_1(t)e^{t}+v_2(t)e^{-t}\)

(b) Solve the system:

\(\displaystyle e^{t}v_1'+e^{-t}v_2'=0\tag{1}\)

\(\displaystyle e^{t}v_1'-e^{-t}v_2'=e^{t}\tag{2}\)

From (1), we find:

\(\displaystyle v_2'=-e^{2t}v_1'\)

And, then substituting into (2), we get:

\(\displaystyle v_1'=\frac{1}{2}\implies v_1=\frac{1}{2}t\)

\(\displaystyle v_2'=-\frac{1}{2}e^{2t}\implies v_2=-\frac{1}{4}e^{2t}\)

Thus:

\(\displaystyle y_p(t)=\frac{1}{2}te^{t}-\frac{1}{4}e^t\)

We may discard the term that simply adds a solution to the homogeneous equation, and state:

\(\displaystyle y_p(t)=\frac{1}{2}te^{t}\)

Because no work was shown, I didn't know until I actually worked the problem that this was the case. :D

 

[shamieh]

We may discard the term that simply adds a solution to the homogeneous equation, and state:

\(\displaystyle y_p(t)=\frac{1}{2}te^{t}\)

Because no work was shown, I didn't know until I actually worked the problem that this was the case. :D

Mark, if my particular solution $y_p = ue^t + ve^{-t}$ then why are you "discarding" the solution for $v$ ? I mean isn't that part of my particular solution? Why would you throw $v$ away? If that was the case then why wouldn't I just set my $y_p$ (particular solution) to $y_p = ue^t$ in the first place?

Thanks for all your help.

 

[MarkFL]

All terms in the particular solution must be linearly independent of all terms in the homogeneous solution. Those that are not may be discarded because they add no useful information.

Adding:

\(\displaystyle -\frac{1}{4}e^t\)

to the general solution when we already have:

\(\displaystyle c_1e^t\)

as one of the terms results in us still just having an arbitrary constant times $e^t$ in the general solution, so we may just discard it.

 

[shamieh]

Oh oh oh, I see. Awesome. That makes sense. It's essentially useless bc we already have it. I get it now. Thanks