Why Is My Differential Equations Homework So Challenging?

[DieCommie]

First let me say I am 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. 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 $$y\dot(0) = -2$$, $$y(0) = -1$$

Homework Equations

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.

 

[Dr Transport]

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.

 

[DieCommie]

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]

As long as the "right hand side" does not involve any functions that also satisfy the homogeneous equation, yes, you can "ignore" the solution to the homogeneous equation.

Find the first and second derivatives of Acos(3x)+ Bsin(3x), plug them into the equation and solve the resulting equations for A and B.

You do NOT want
$$\alpha e^5t + \beta e^-4t = Acos(3t) + B sin(3t)$$