One of the solutions of $y^{(3)}+2y''-16y=0$ is $y=e^{2x}$. Find the general solution.
I was thinking of finding the characteristic equation and solving, but I feel like there may be an easier way to approach this. Any help?
Thank you!
2 Answers
I assume this is homework, so let's just recall the general theory. You're supposed to write down the characteristic (or auxiliary) equation, which in this case is $$ r^3 + 2r^2 -16 = 0$$ There will be three solutions to this, given by $r_1, r_2, r_3$, and if they are all distinct then we know that the general solution is given by a linear combination of the functions $e^{r_1 x}$, $e^{r_2 x}$, and $e^{r_3 x}$, and if there are repeated roots you also multiply additional solutions by powers of $x$.
So all that one would need to do to solve your problem in general is to factor $r^3 + 2r^2 - 16$. In general, factoring cubics is pretty hard, which is why the problem gives you that $e^{2x}$ is a solution, which means that $(r-2)$ is one of the factors of the above cubic polynomial. By factoring it out you get a quadratic, from which you can obtain the remaining two roots.
Here's one silly way to do it. Use reduction of roots. One solution is $y = e^{2x}$. Guess $w = yv$. Then, $$w' = yv' + y'v$$ $$ w'' = yv''+2y'v' + y''v$$ $$w''' = yv'''+3y'v''+3y''v'+vy''' $$
Plugging in $w$ gives after simplifying $$e^{2x}(v'''+8v''+20v') = 0$$
Solving gives (according to WA, since I'm too lazy to do it by hand) $$v = \frac{1}{10}e^{-4x}((c_2-2c_1)\sin(2x)-(c_1+2c_2)\cos(2x)) + c_3$$ Then multiplying by $y$ gives your answer.
