# Differential Equation - SOLDE

First let me say Im very frustrated at this and could use some real through advice. My school says to not take differential equations for a BS in physics. They instead give us a real poor packet to learn it on our own with 'guidance' from a teacher. :uhh: I am wishing I took diff. eq. anyway...

## Homework Statement

$$y\ddot - y\dot -20 y = 17sin(3t)$$

1) Find the general soltuon for homogenous
2)find particular soltuons
3)fin solution for $$y\dot(0) = -2$$, $$y(0) = -1$$

## The Attempt at a Solution

First I need to solve the homo. part. I think I can do that.
I find the characteristic equation to be $$r^2 -r -20 = 0$$ and get $$y(t) = \alpha e^(2t) + \beta e^(-t)$$

I now need to find a particular solution. I have no idea how to do this. A table in my 'packet' says for inhomogeneity of C sin(wt) , the general form of $$y_p(t)$$ is $$A cos (\omega t) + B sin (\omega t)$$.

What do I do with this $$y_p(t)$$ thing? Please give me some very detailed hints on this part! Thank you.

If I can get the particular solution I think I can apply the boundary conditions.

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Dr Transport
Gold Member
Check your homogeneous solutionagain, you may have typed in a mistake. As for the particular solution, $$\omega = 3$$ so plug in your solution and match coefficients to obtain a final soluiton.

Check your homogeneous solutionagain, you may have typed in a mistake. As for the particular solution, $$\omega = 3$$ so plug in your solution and match coefficients to obtain a final soluiton.
Ok I found small mistake and now have the correct homo. soltution. $$y(t) = \alpha e^(5t) + \beta e^(-4t)$$.

I am still quite lost as what to do with the particular solution...

I have to add it to the homo. solution? Do I just ignore the homo. solution? I know omega = 3, you say 'so plug in your solution and math coefficents' Do set the particular soltion equal to the homo. solution? What coefficents do I match...
I this what you mean ?$$\alpha e^5t + \beta e^-4t = Acos(3t) + B sin(3t)$$

HallsofIvy
$$\alpha e^5t + \beta e^-4t = Acos(3t) + B sin(3t)$$