# Problem with Complex Second Order Equations

1. Feb 13, 2005

### Lyuokdea

Ok, this one is really sticking me up:

$$y'' - 2y' + 2y = e^{t}cos(t)$$

I solved the homogenous version and got roots 1 +/- i and put these in to get the equation

$$y_h = c_1e^tcos(t) + c_1e^tsin(t)$$

And I found that the root for e^tcos(t) should be (D- (1 +/- i)

But I'm completely stuck on what to do after this, the part that really confuses me is that in the answer, they only have a e^t sin(t) factor in the answer in addition to the normal homogenous answer, why does sin(t) appear without cos(t) ?

~Lyuokdea

2. Feb 13, 2005

### dextercioby

Did u try to apply Lagrange's method...?

Daniel.

3. Feb 13, 2005

### Lyuokdea

I don't believe so, I don't think we were ever taught about that, is there another way to solve it? Maybe we called it something different. I'll go back and look through my notes

~Lyuokdea

4. Feb 13, 2005

### dextercioby

Search for "Method of Variation of Constants" or "Method of Lagrange"...

Daniel.

5. Feb 13, 2005

### Lyuokdea

ok, I got it, thanks

6. Feb 15, 2005

### HallsofIvy

Staff Emeritus
You should be able to do this by basic "undetermined coefficients".

Normally, if you have "right hand side" like etcos(t), you would try
y= et(Acos(t)+ Bsin(t)).

However, here, functions of that form already satisfy the homogenous equation and can't give you anything but 0 on the right side.

Okay, so you multiply by t:
try y= ett(Acos(t)+ Bsin(t)).

The rest is cranking away.

7. Feb 15, 2005

### saltydog

You guys mind if I finish up? He's gone to something else I think. You know school and all:

$$y'' - 2y' + 2y = e^{t}cos(t)$$

Because the RHS is a particular solution of a linear ODE, namely:

$$(D^2-2D+2)y=0$$

can apply this operator to both side of non-homogeneous eq. to collapse the RHS:

$$(D^2-2D+2)(D^2-2D+2)y=0$$

Since the solution to this equation is:

$$y=c_1e^x\cos(x)+c_2e^x\sin(x)+Axe^x\sin(x)+Bxe^x\cos(x) [/itex] then the particular solution to the non-homogeneous equation must be of the form: [tex]y_p=Axe^x\sin(x)+Bxe^x\cos(x)$$

Substituting this equation into the non-homogeneous equation and equating coefficients, we find B=0 and A=1/2.

Thus the general solution is:

$$y(x)=c_1e^x\cos(x)+c_2e^x\sin(x)+\frac{1}{2}xe^x\sin(x)$$

Using $c_1=1$ and $c_2=0$, I plotted a graph of the solution which is attached.

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