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Suppose $y''-4y=xe^{2x}$.

Solution of the homogenous equation is $y_H(x) = C_{1} e^{2x} + C_{2} e^{-2}$, after solving the characteristic equation with the guess $y=e^{rx}$.

Now my instructor insists that the one solution is either

$y(x) = Ax^{2} e^{2x} + Bxe^{2x}$

or

$y(x) = Ax^{2}e^{2x} + Bxe^{2x}+Ce^{2x}$

...and then I get the solution $y(x) = y_{H}(x) +\frac{1}{8}x^{2}e^{2x}-\frac{1}{16}xe^{2x}$

but I cannot understand the choices. Why do just the forms work? I have tried many different forms there and get lost/wasted a lot time -- and my instructor just says that it is a "guess" and smiles. How can I find the one solution without guessing? For one solution, things such as $y(x)=e^{2x}$ or $y(x)=xe^{2x}$ work so I am a bit lost why the more complex form is "the solution" and why it provides all solutions.

For example, the case $y=e^{2x}$. We get $y'=2e^{2x}$ and $y''=4e^{2x}$ so

$y''-4y = 4e^{2x} - 4 e^{2x} = 0$

so $RHS = xe^{2x} = 0$ so $x=0$, a solution?! If not why? Ok it is naive, it does not contain all cases but how can I know that some form contains all?

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At the "guess" level, $Bxe^{2x}+Ce^{2x}$ is very reasonable. And it would work nicely, with anything other than $-4y$, like $-3y$, or $7y$. But substituting the "guess" in $y''-4y$ shows that the $-4y$ makes the $xe^{2x}$ term disappear. Now maybe one realizes that if you add $Ax^2e^{2x}$ to the guess, and calculate $y''-4y$, the $x^2e^{2x}$ term will similarly disappear, which now is a good thing! (As you can imagine, there is also general theory available, but playing around and paying attention to structure works well in this case.) –  André Nicolas May 14 '11 at 19:47
    
@ahh: What is now at the end of the post is not right, and seems to show a misunderstanding of the process. You want a "guess" $y$ such that $y'' -4y$ is identically equal to $xe^{2x}$, that is, equal for all $x$. The "solving" for $x$ that you did is totally irrelevant, has nothing to do with the question. You want a "guess" $y$ such that $y'' -4y$, when simplified, is $xe^{2x}$. –  André Nicolas May 14 '11 at 19:56
    
@user6312: sorry I don't follow. (x^{2}e^{2x})' = 2xe^{2x} + 2x^{2}e^{2x}$ and $(x^{2}e^{2x}) = 6xe^{2x} + 4x^{2}e^{2x} + 2e^{2x}$ -- I get just more terms, nothing really disappearing?! If there is, how can you see it so quickly? –  hhh May 14 '11 at 19:58
    
the solution to your non-homogeneous equation should be looked on those of the form $Ap(x)e^{rx}$ if $r$ is not a solution to your characteristic polynomial. Unfortunately, 2 is; so you should look for a solution of the form $Ax^qp(x)e^{rx}$, where $q$ is the multiplicity of your root r. (Adding of course the one to the homogeneous equation) –  Andy May 14 '11 at 21:12
    
755 views but only one up vote? –  user66360 Dec 15 '13 at 17:40

4 Answers 4

up vote 2 down vote accepted

Another way to do this is using the Exponential Shift Theorem (see e.g. http://www.math.ubc.ca/~israel/m215/coco/coco.html): For any polynomial $P$, constant $k$ and function $u$, $P(D) e^{kx} u = \exp(k x) P(D+k) u$ (where $D$ stands for derivative). In your case $P(t) = t^2 - 4$, $k=2$, and $P(t+2) = (t+2)^2 - 4 = t^2 + 4 t$ so if $y = e^{2x} u$ we have $y'' - 4 y = e^{2x} (u'' + 4 u')$. Now you want $y'' - 4 y = x e^{2x}$ so $u'' + 4 u' = x$. Writing $v = u'$, we have the first-order linear equation in $v$: $v' + 4 v = x$, which has a solution $v = \frac{x}{4} - \frac{1}{16}$. An antiderivative of this is $u = \frac{x^2}{8} - \frac{x}{16}$, corresponding to the particular solution $y = e^{2x} \left(\frac{x^2}{8} - \frac{x}{16}\right)$ of your original equation.

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I am stuck to this point in the example on the site "We have $y=e^{-x}u$. We have $P(D-1)u = D^{2}u = 4$, so one solution is $u=2x^{2}$.", not evident to me. When I calculated it open, I got $(D+1)^{2}=4e^{-x} \rightarrow (D+1)^{2}e^{x}=4$ and then by the theorem, let $P(D)=(D+1)^{2}$, $P(D)e^{x}=e^{x}P(D+1)=e^{x}(D+2)^{2}$ but $P(D)e^{x} = 4$. Did I do s/thing wrong or can I still get to the general solution? –  hhh May 15 '11 at 17:25
1  
The example was $y'' + 2 y' + y = 4 e^{-x}$. Thus $P(t) = t^2 + 2 t + 1 = (t+1)^2$. Since $e^{-x}$ is on the right side, we take $k=-1$. Now $P(t-1) = (t-1+1)^2 = t^2$. So take $y = e^{-x} u$, and the theorem says $P(D) y = e^{-x} P(D-1) u = e^{-x} D^2 u$. We want this to be $4 e^{-x}$, so $D^2 u = 4$. –  Robert Israel May 15 '11 at 18:42
    
When the solutions $xe^{-x}$, $e^{-x}$ and $2x^{2}e^{-x}$ are known, how do you know that the general solution is just their sum $y_{G} = 2C_{1}x^{2}e^{-x}+C_{2}xe^{-x}+C_{3}e^{-x}$? –  hhh May 15 '11 at 19:33
1  
From $D^2 u = 4$, integrate once to get $D u = 4 x + C_1$, then integrate again to get $u = 2 x^2 + C_1 x + C_0$, where $C_1$ and $C_0$ are arbitrary constants. –  Robert Israel May 16 '11 at 4:11

The "right" approach is highly context-dependent. What follows is from the perspective of an introduction to differential equations attached to a first-year calculus course.

We describe an informal procedure, in which we make a plausible guess as to $y$, and test how well the guess works by calculating $y''-4y$ for this guess. We then use the information obtained from the calculation of $y''-4y$ to improve our guess. Of course more formal procedures are available, but it is nice to see how far one can get by playing a little.

A very reasonable "guess" for a particular solution is $y=xe^{2x}$. Let's check whether it works. If $y=xe^{2x}$, then $y'=2xe^{2x}+e^{2x}$, and $y''=4xe^{2x}+4e^{2x}$. Then $$y''-4y=4e^{2x}$$

But $4e^{2x}$ is definitely not the same function as $xe^{2x}$. And multiplying the guess $xe^{2x}$ by a constant $k$ will do nothing useful.

Note that things would have worked out nicely if the differential equation had, for example, $y''-3y$ on the left instead of $y''-4y$. But the $-4y$ is exactly the right thing to make the $xe^{2x}$ term disappear. When the person posing the problem chose $-4$, (s)he chose the only constant that would make our lives difficult. Nasty!

However, this gives us an idea. If we try $y=x^2e^{2x}$, and calculate $y''-4y$, maybe the $x^2e^{2x}$ term will disappear, which is exactly what we want. Calculate. If $y=x^2e^{2x}$, then $y'=2x^2e^{2x}+ 2xe^{2x}$ and $y''=4x^2e^{2x}+8xe^{2x} +2e^{2x}$. Thus $$y''-4y=8xe^{2x} +2e^{2x}$$

Close! We should clearly multiply our guess by $(1/8)$ to get the $xe^{2x}$ term right. And we will not be quite there, since then there will still be an unwanted $(1/4)e^{2x}$ term in $y''-4y$ to get rid of.

Look back on the work we did at the beginning with $y=xe^{2x}$. Then $y''-4y$ turned out to be $4e^{2x}$. So to get rid of the unwanted $(1/4)e^{2x}$, we can simply add $(-1/16)xe^{2x}$ to the "guess" $(1/8)x^2e^{2x}$.

So we end up with the particular solution $$y =\frac{x^2e^{2x}}{8}-\frac{xe^{2x}}{16}$$

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Perhaps it would help to see the general method.

You want to solve an equation $Ly=x^n e^{\lambda x}$, where $L$ is a constant linear differential operator - in your case $Ly=y''-4y$. Let us solve $Ly=e^{(\lambda+\epsilon) x}$ instead: if we substitute $y=c(\epsilon)e^{(\lambda+\epsilon) x}$, we get $c(\epsilon)p(\lambda+\epsilon)=1$ (where $p$ is the characteristic polynomial of $L$, in your case $p(t)=t^2-4$), i.e. $c(\epsilon)=1/p(\lambda+\epsilon)$. We now expand $$L\frac{e^{(\lambda+\epsilon) x}}{p(\lambda+\epsilon)}=e^{(\lambda+\epsilon) x}$$ to a power series in $\epsilon$, look at the term at $\epsilon^n$, and get a solution of $Ly=x^n e^{\lambda x}/n!$.

In your case $\lambda=2$, $p(t)=t^2-4$, and $c(\epsilon)=1/(\epsilon(4+\epsilon))$. We thus get

$$L\bigl((\epsilon^{-1}4^{-1}+(4^{-1}x-4^{-2})+\epsilon(4^{-1}x^2/2-4^{-2}x+4^{-3}) +\dots )e^{2 x}\bigr)=$$ $$=(1+\epsilon x+\epsilon^2 x^2/2 +\dots)e^{2 x}.$$ Looking at the term at $\epsilon$, we get $$L((4^{-1}x^2/2-4^{-2}x+4^{-3})e^{2 x})=xe^{2 x},$$ i.e. $y=(4^{-1}x^2/2-4^{-2}x+4^{-3})e^{2 x}$ is a solution of your equation (we can drop $4^{-3}e^{2 x}$ from the solution, as it solves the homogeneous equation).

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I believe this can be done without guesswork.

One way to get this mechanically is to use the method of Laplace Transforms.

Of course, I haven't tried out it out myself, but given that

$$\mathcal{L}[t^n e^{at}] = \frac{n!}{(s-a)^{n+1}}$$

I believe the approach will work for your problem.

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...but does that return all solutions if you take the laplace transform and then the inverse? –  hhh May 15 '11 at 10:20
    
@hhh: I believe it does. Why do you think it might not? –  Aryabhata May 15 '11 at 14:31

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