Problem with Complex Second Order Equations

[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 into 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) ?

Thanks for your help,

~Lyuokdea

 

[dextercioby]

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

Daniel.

 

[Lyuokdea]

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

Daniel.

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

 

[dextercioby]

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

Daniel.

 

[Lyuokdea]

ok, I got it, thanks

 

[HallsofIvy]

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

Normally, if you have "right hand side" like e^tcos(t), you would try
y= e^t(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= e^tt(Acos(t)+ Bsin(t)).

The rest is cranking away.

 

[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:

[tex]y=c_1e^x\cos(x)+c_2e^x\sin(x)+Axe^x\sin(x)+Bxe^x\cos(x) [/itex]<br /> <br /> then the particular solution to the non-homogeneous equation must be of the form:<br /> <br /> $$y_p=Axe^x\sin(x)+Bxe^x\cos(x)$$<br /> <br /> Substituting this equation into the non-homogeneous equation and equating coefficients, we find B=0 and A=1/2.<br /> <br /> Thus the general solution is:<br /> <br /> $$y(x)=c_1e^x\cos(x)+c_2e^x\sin(x)+\frac{1}{2}xe^x\sin(x)$$<br /> <br /> Using $c_1=1$ and $c_2=0$, I plotted a graph of the solution which is attached.[/tex]