Toronto Math Forum
MAT2442013S => MAT244 MathTests => Term Test 1 => Topic started by: Victor Ivrii on February 13, 2013, 10:39:58 PM

Find the general solution for equation
\begin{equation*}
y'' + 4y'+5y =t e^{2t}+ e^{2t}\cos(t).
\end{equation*}

Funny story: after I posted this I also checked my work against wolfram alpha (http://www.wolframalpha.com/input/?i=y%27%27+%2B+4y%27+%2B+5y+%3D+t+e^%282t%29+%2B+e^%282t%29+cos%28t%29), which said I got it wrong for the homogeneous part. I spent some time trying to spot any error, but there isn't any. I am smarter than the machine!
(You can press "show step by step solution" and wolfram alpha will come up with what I came up with, so that's why I'm so confident. I filed a bug report, maybe someday this will be fixed...)
$\renewcommand{Re}{\operatorname{Re}}$
Here we use the method of undetermined coefficients. First we find the characteristic equation and its derivative (which will definitely be useful):
$$Q(r) = r^2 + 4r + 5$$
$$Q^\prime(r) = 2r + 4$$
Now we solve for $te^{2t}$, for which our exponent is $r_1 = 2$.
$$Q(r_1) = 4  8 + 5 = 1$$
$$Q^\prime(r_1) = 4 + 4 = 0$$
So we want to use $t e^{2t}$. If we plug that in we get $L[t e^{2t}] = e^{2t}(t Q(2) + Q^\prime(2)) = t e^{2t}$. This is what we want, so we're done the first part.
Now we solve for $e^{2t} cos(t)$, for which the exponent is $r_2 = 2 + i$.
$$Q(r_2) = (2+i)^2 + 4(2 + i) +5 = 4  4i  1  8 + 4i + 5 = 0$$
$$Q^\prime(r_2) = 2(2+i) + 4 = 4 + 2i + 4 = 2i$$
So we want to use $t e^{(2 + i)t}$. If we plug that in, we get $L[t e^{(2 + i)t}] = e^{(2 + i)t}(t Q(2 + i) + Q^\prime(2 + i)) = (2i)e^{(2 + i)t}$.
We don't want that 2i, so let's divide it out in the input. It turns out that $1/2i = i/2 = i/2$. So $L[\frac{i}{2} t e^{(2 + i)t}] = e^{(2 + i)t}$, which is nearly right.
Note:
$$\Re e^{(2 + i)t}) = \Re (e^{2t}e^{it}) = \Re (e^{2t}(\cos t + i \sin t)) = e^{2t} \cos t$$.
Since $L$ is linear, if we take only the real part of the input, we'll get only the real part of the output. So we need to compute the real part of $\frac{i}{2} t e^{(2 + i)t}$.
To do that let's multiply it out:
$$\frac{i}{2} t e^{(2 + i)t}$$
$$\frac{i}{2} t e^{2t}(\cos t + i \sin t)$$
$$t e^{2t}(\frac{i}{2}\cos t + \frac{i}{2}i \sin t)$$
$$t e^{2t}(\frac{i}{2}\cos t + \frac{1}{2}\sin t)$$
The real part of that is $\frac{1}{2}t e^{2t}\sin t$, so that $L[\frac{1}{2}t e^{2t}\sin t] = e^{2t} \cos t$.
So then we put our two subproblem solutions together and we get a particular solution, thanks to the linearity of $L$. Our particular solution is:
$$Y(t) = t e^{2t} + \frac{1}{2}t e^{2t}\sin t$$
What remains is to compute a general solution for the homogenous part, and we can combine that to find the general solution for the nonhomogenous ODE.
During the above work we found that $2 + i$ was a root of the characteristic equation. So the conjugate is also a root, $2  i$, and we can take the real and imaginary parts of $e^{2 \pm i}$ to find the solution, which is:
$$y_{gen[homogeneous]}(t) = c_1 e^{2t}\cos(t) + c_2 e^{2t}\sin (t)$$
So the general solution for the nonhomogeneous ODE is:
$$y(t) = c_1 e^{2t}\cos(t) + c_2 e^{2t}\sin (t) + t e^{2t} + \frac{1}{2}t e^{2t}\sin (t)$$

my solution

Devin, for me WolframAlpha gave exactly your answer for homogeneous equation. Probably you asked it not politely enough :D
PS Usage of double dollars in LaTeX is deprecated. There is command \Re but out of the box it returns $\mathfrak{R}$ so I redefined it
$\renewcommand{\Re}{\operatorname{Re}}$
(dollars needed to tell MathJax to pay attention, in normal LaTeX they would be wrong)
Marcia, I decided that variation of parameters deserves a reward. Note however that you made small computational errors

for me (http://www.wolframalpha.com/input/?i=y%27%27+%2B+4y%27+%2B+5y+%3D+t+e^%282t%29+%2B+e^%282t%29+cos%28t%29), wolfram alpha spits out $y(t) = c_1 e^{2 t} \sin(t)+c_2 e^{2 t} \cos(t)+\frac{1}{2} e^{2 t} (t (\sin(t)+2)+\cos(t))$. It's technically correct, because you can just change the constants to turn it into my solution, but...
PS Usage of double dollars in LaTeX is deprecated.
For most everything what I really wanted was to use "align*". Turns out mathjax supports this, so I'll be more latexy in the future.
Thanks for the advice!