# Homework Help: Differential Equation

1. May 17, 2014

### Calu

1. The problem statement, all variables and given/known data

I have a differential equation: $\ddot{x} -2\dot{x} + 5x= 10 + 13cos(3t)$

2. Relevant equations

x(t) = xc + xp
where xc is the Complementary Function and xp is the Particular Integral.

3. The attempt at a solution

I have formed and solved the auxiliary equation:

$m^{2} - 2m + 5 = 0$

$m_{1} = 1 + 2i , m_{2} = 1 - 2i$

How would I go about forming xc = et(Acos2t + Bsin2t) ?

Furthermore, how would I choose a trial function to determine the Particular Integral?

Last edited: May 17, 2014
2. May 17, 2014

### LCKurtz

That's "complementary".

I don't understand that question. You did just "form" it, didn't you?

Well, since you need a constant and a $\cos t$ after you differentiate your $x_p$ I would suggest something of the form $x_p = A + B\cos t + C\sin t$.

3. May 17, 2014

### Calu

I give up with English. Fixed the OP.

We have taught to from what is called an auxiliary function by taking the coeffeicents of $\ddot{x}, \dot{x},$and $x$ and (say a, b, c) and forming the equation

$am^{2} + bm + c$

and then solving this equation to find 2 values of m (m1 and m2) which are to then form the basis of the complementary function xc. The complementary function I've given above is what I have to form given my values m1 and m2.

That makes sense, thanks. In the given answer a constant 3 has been added so that it reads:

$x_p = A + B\cos 3t + C\sin 3t$. Is there any particular reason that has been chosen?

4. May 17, 2014

### LCKurtz

Your original post had a $\cos t$ on the right side. Was that a typing error and it was supposed to be $\cos(3t)$? That would be the only reason to use $3t$ in the trig functions for your $x_p$.

5. May 17, 2014

### Calu

Ahh yes, my mistake. Thank you.

Do you have any idea about forming the complementary function?

6. May 17, 2014

### Staff: Mentor

You already have the complementary function (i.e., the solution to the homogeneous problem). Are you asking about how to write a particular solution (not the complementary solution)? (BTW, the p subscript in xp stands for "particular.")

If not, then I don't understand what you're having trouble with.

7. May 17, 2014

### Calu

I don't know how to go from having m1 and m2 to making the equation for xc. The equation in my post has been given to me in the answer, I don't know how to get it myself.

8. May 17, 2014

### LCKurtz

Your characteristic equation comes by looking for a solution $x=e^{mt}$ and you have found that $m = 1\pm 2i$. That means the general solution of the homogeneous equation is$$x_c = Ae^{(1+2i)t} + Be^{(1-2i)t}= e^{t}(Ae^{2it} + Be^{-2it})$$Now use the Euler formulas for the complex exponentials in terms of sines and cosines to get it in the {sine,cosine} form.

9. May 18, 2014

### Calu

$x_c = Ae^{(1+2i)t} + Be^{(1-2i)t}= e^{t}(Ae^{2it} + Be^{-2it})$

I did this, but found that I get:

$x_c = e^{t}((A + B)cos2t + (A - B)isin2t)$

Can these arbitrary constants be combined to yield something like:

$x_c = e^{t}(Ccos2t + Disin2t)$ for some arbitrary constants C, D ?

Last edited: May 18, 2014
10. May 18, 2014

### HallsofIvy

Yes, of course. Since A and B are arbitrary constants, so are A+ B and A- B.

11. May 18, 2014

### Calu

Thanks.

One more question. Is the $i$ term meant to be present here:

$x_c = e^{t}(Ccos2t + Disin2t)$ ?

12. May 18, 2014

### LCKurtz

Include the $i$ in the $D$. While it appears the constants C and D may be complex, if the DE has real coefficients and real boundary conditions, they will come out real. If you left your answer in the $\{e^{2it},e^{-2it}\}$ form the constants would come out complex.

13. May 18, 2014

### Calu

Oh I see, thanks a lot!