Second degree linear derivate with exponential function, which general solution?

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$is the multiplicity of your root r. (Adding of course the one to the homogeneous equation) – Andy May 14 '11 at 21:12 755 views but only one up vote? – user66360 Dec 15 '13 at 17:40 4 Answers Another way to do this is using the Exponential Shift Theorem (see e.g. http://www.math.ubc.ca/~israel/m215/coco/coco.html): For any polynomial$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)$of your original equation. - I am stuck to this point in the example on the site "We have$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$. Did I do s/thing wrong or can I still get to the general solution? – hhh May 15 '11 at 17:25 The example was$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 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 $$y =\frac{x^2e^{2x}}{8}-\frac{xe^{2x}}{16}$$ - 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