# Finding the particular solution to a trigonometric differential equation? [closed]

This problem that I'm linking is the first in a long list of homework problems where we find the particular, complementary, and total solutions to differential equations. I'm familiar with how to find the particular solution when it comes to linear and exponential differential equations, but I'm not so sure when a trigonometric function is involved. Is there any way I could get a step by step guide on how to do part two of the below homework problem? Thank you.

Problem: https://imgur.com/8Ic74RH

## closed as off-topic by JJacquelin, José Carlos Santos, steven gregory, user1729, AbcdMar 18 at 20:06

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• I vote to close your question because I cannot open the link, so the wording of the problem is not accessible. – JJacquelin Mar 18 at 8:02

Consider the general problem, $$y' + my=A \cos (\omega t + \phi)u(t)$$ This is an ordinary LDE of the first order, of the form, $$y' + Py=Q$$ Then, the general solution is, $$ye^{mt}=A\int e^{mt} \cos (\omega t + \phi)u(t) dt + C$$ Where $$C$$ is the constant of integration. For, $$u(t)$$ is unknown we can say nothing further. If $$u(t)=B$$ were constant, we could write, $$y=\frac {AB}{m^2+\omega^2} [\omega \sin (\omega t + \phi)+m \cos (\omega t + \phi)] + Ce^{-mt}$$
• I believe $u(t)$ is the Heaviside function. The OP needs to clarify this. – Sean Roberson Mar 18 at 2:48