Second-order ODE problems

  • #1
There are a couple of problems on this week's homework assignment that are giving me trouble.

(1) Find a particular solution for [itex]y'' + y = t^2[/itex] by using the method of undetermined coefficients

Here I don't know which coefficient expression to use, for example if the term on the right side was e^t, I could sub y = Ae^t so that all 'e^t's would cancel out and I would be left with an expression for A.

(2) Solve the IVP: [itex]y'' - 4y' +2y = e^{2t}[/itex], homogenous initial conditions at t=0.

What I did was what I normally do for any first-order ODE. I separated the problem out into homogenous and particular parts.
[tex] y_h: y''-4y'+2y=0[/tex]
[tex]s^2-4s+2=0\rightarrow s=2\pm\sqrt{2} [/tex] where s is a characteristic root. Therefore [itex]y_h=c_1 e^{(2+\sqrt{2})t}+c_2 e^{(2-\sqrt{2})t}[/itex]
For the particular part I used undetermined coefficients and subbed y=Ae^2t and got an expression for A: 4A - 8A + 2A = 1, so A = -1/2, and [itex]y_p = -\frac{1}{2}e^{2t}[/itex]
What I am confused about is what comes next.
Is the solution then just [itex]y = y_p + y_h = c_1 e^{(2+\sqrt{2})t}+c_2 e^{(2-\sqrt{2})t} -\frac{1}{2}e^{2t}[/itex] ? And how do I account for the initial conditions?
 

Answers and Replies

  • #2
dextercioby
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Initial conditions help you fix the coefficients values.They are arbitrary,unless there are initial conditions.

For the first,substitute

[tex] y_{p}(t)=C(x) \sin x [/tex]

Daniel.
 

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