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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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    755 views but only one up vote?2013-12-15

4 Answers 4

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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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    @hhh: I believe it does. Why do you think it might not?2011-05-15
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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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    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.2011-05-16
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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 $y =\frac{x^2e^{2x}}{8}-\frac{xe^{2x}}{16}$