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$y''+y = e^{it}+e^{3it}$
Solution
$y = Ae^{it}-\dfrac{1}{8}e^{3it}-\dfrac{it}{2}e^{it}$
This seems correct. See this.$y''+4y=1+\sin t+\sin 2t$
Solution
$y=A\cos 2t + B\sin 2t + \dfrac{1}{4} + \dfrac{1}{3}\sin t - \dfrac{t}{4}\cos 2t$
Correct?
When you solve inhomogeneous linear constant coefficient ODE's you start with the general solution to the homogeneous equation, which in this case is:$y''+y = e^{it}+e^{3it}$
Solution
$y = Ae^{it}-\dfrac{1}{8}e^{3it}-\dfrac{it}{2}e^{it}$
Hi dwsmith,If we re-write the problem, we have
\[y''+y=\cos t + \sin t + e^{3it}\]
That is the same thing as:Online Mathematica says the solution is $y=A\cos t + B\sin t -\dfrac{it}{2}e^{it} + \dfrac{1}{4}e^{it} - \dfrac{1}{8}e^{3it}$.
The coefficient \(c_{1}\) is incorrect. It should be, \(c_{1}=-\dfrac{i}{2}\). Then you will get,I don't see how the term with the 1/4 appears at all.
Writing the equation as $y''+y = \cos t + i\sin t + e^{3it}$ didn't produce anything promising.
Then $y_p = c_1t\cos t + c_2it\sin t + c_3e^{3it}$ so $y''_p = -2c_1\sin t - c_1t\cos t +2c_2i\cos t - c_2it\sin t -9c_3e^{3it}$.
$y''_p+y_p = 2c_2i\cos t - 2c_1\sin t - 8c_3e^{3it}$
So $c_1 = -1/2$, $c_2 = -i/2$, and $c_3 = -1/8$.
$$
y_p = -1/2t\cos t + 1/2t\sin t -1/8e^{3it}
$$
So the only thing wrong with my original solution was that it was missing $Be^{-it}$ then. Why didn't you all say that to begin with?The coefficient \(c_{1}\) is incorrect. It should be, \(c_{1}=-\dfrac{i}{2}\). Then you will get,
\[y_p = -\frac{it}{2}\cos t + \frac{t}{2}\sin t -\frac{1}{8}e^{3it}\]
\[y_p = -\frac{it}{2}e^{it} -\frac{1}{8}e^{3it}\]
The general solution is,
\[y=Ce^{it}+De^{-it}-\frac{1}{8}e^{3it}-\frac{it}{2}e^{it}\]
This can be written as,
\[y=\left(C-\frac{1}{4}+\frac{1}{4}\right)e^{it}+De^{-it}-\frac{1}{8}e^{3it}-\frac{it}{2}e^{it}\]
\[\therefore y=\left(C-\frac{1}{4}\right)e^{it}+De^{-it}+\frac{1}{4}e^{it}-\frac{1}{8}e^{3it}-\frac{it}{2}e^{it}\]
Let \(\displaystyle E=\left(C-\frac{1}{4}\right)\). Then,
\[y=Ee^{it}+De^{-it}+\frac{1}{4}e^{it}-\frac{1}{8}e^{3it}-\frac{it}{2}e^{it}\]
Does this clarify your doubts?![]()