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$f(x)=Q(x)e^{kx}\cos(mx)+P(x)e^{kx}\sin(mx)$

http://www.math.pitt.edu/~athanas/MATH-0230-CALCULUS-II/notes-diffeq.pdf

you can see the case II. I cannot understand this part..

then what will be the partial solution of this equation? please explain.

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    I don't see a differential equation. – 2017-01-27
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    please think of it as a n-th order problem. thank you for your understading. – 2017-01-27
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    This is very unclear. Are you looking to solve for $f(x)$? If yes, then just integrate twice. – 2017-01-27
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    http://www.math.pitt.edu/~athanas/MATH-0230-CALCULUS-II/notes-diffeq.pdf please take a look at the case II. please help me. – 2017-01-27
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    In the link, that isn't $f(x)$ but $y_p(x)$, which is the partial solution to a differential equation of the form $ay''+by'+cy=f(x)$ where $f(x)=P(x)e^{kx}\sin(mx)$ or $f(x)=P(x)e^{kx}\cos(mx)$. So that $is$ the partial solution. Was that your question? – 2017-01-27
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    if $f(x)=Q(x)e^{kx}cos(mx)+P(x)e^{kx}sin(mx)$ then what will be my solution? – 2017-01-27
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    "If f(x) is a sum of terms, like $f(x)=x^2+e^{-x}+\cos(x)$, do it as separate problems solving for $y_{p1}, y_{p2}$". So you just proceed to look for the coefficients in the same fashion you would do if $f=Pe^{kx}\cos(x)$ (also with $\sin(x)$, and just sum them – 2017-01-27

1 Answers 1

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In full answer form, if $ay''+by'+cy=\underset{f1}{\underbrace{P(x)e^{kx}\sin(mx)}}+\underset{f_2}{\underbrace{Q(x)e^{kx}\cos(mx)}}$, then you are to set, for $f_1$ : $y_1=P_1(x)e^{kx}\sin(mx)+P_2(x)e^{kx}\cos(mx)$, where $P_1$ and $P_2$ are of the same degree as $P(x)$, and solve for the coefficients of $P_1$ and $P_2$.

The same for $f_2$ : $y_2=Q_1(x)e^{kx}\sin(mx)+Q_2(x)e^{kx}\cos(mx)$.

The result is then $y=y_1+y_2+y_c$ where $y_c$ is the solution to the homogenous DE.