# Second Order DE - Need Help With Equation

1. Oct 25, 2009

### Clef

Hey Guys,

I've been banging my head with this one for a few days now. I've done (a) and (b) but I'm yet to do (c) if someone could complete this for me or help me along the way that would be greatly appreciated.

Thanks heaps!

2. Oct 25, 2009

### l'Hôpital

Do you know anything about solving constant-coefficient second order DEs?

Suggestion: Solve the homogemous, find a particular solution, then i = i_t + i_p where i_t = Solution to teh homogenous, and i_p = particular solution.

Alternatively, if you knew Laplace Transform, you could try that as well.

3. Oct 25, 2009

### Clef

Okay
I've found my homogenous solution to be
(e^-3t)(Acos4t + Bsin4t)

I'm now attempting the particular solution by changing the original equation into the complex form:

z'' + 6z' +25z = -292e^(i4t)

And solving that. To be honest I am completely stuck where to go from here and my attempts have been going in a very roundabout way.
:(

4. Oct 25, 2009

### elibj123

read some about the variation of parameters method
http://en.wikipedia.org/wiki/Method_of_variation_of_parameters

basically if {u1,u2} are the basis of the homogenous solution space, you assume the private solution
of the NH equation is:
A(t)*u1(t)+B(t)*u2(t)

playing around with the ODE you get the linear system

u1*(dA/dt)+u2*(dB/dt)=0
(du1/dt)*(dA/dt)+(du2/dt)*(dB/dt)=q(t)

where q(t) is the NH part. you get A' and B' and then by integration you get A(t),B(t).
Notice that A*u1+B*u2 will be the general solution for the ODE because A,B will contain
the free parameters (due to integration).

5. Oct 25, 2009

### Clef

I'm really sorry. But I do not completely understand this method. Would it be the same as substituting
y= tAcos4t+tbsin4t into the equation and then equating to get the PS?

6. Oct 25, 2009

### Clef

or is it sufficient to use
i= Acos4t + Bcos4t

for the solution for the particular solution of the whole DE?

Because I read somewhere that for some reason you have to put

i= t(Acos4t + Bsin4t)

but this is very very confusing :'(

7. Oct 25, 2009

### elibj123

I'm familiar with the substitution method but i don't like it because as you said it has many rules depending on the characteristic polynomial of the equation (i am not sure if this method even works for non-linear ode).

Variation of Parameters method is pretty simple once you have the basis of the solution-space.
The technique is very simple (without going into developing the method):
if you have a basis $$u_{j}$$ (in this case you found 2 functions)
then you get n linear equations of n variable $$k^{'}_{j}$$.

each equation is of the form:
$$\sum^{n}_{j=1}(\frac{d^{i}u_{j}}{dt^{i}}*k^{'}_{j})=0, \forall i=0,...,n-2$$

and for i=n-1 you get the equation:
$$\sum^{n}_{j=1}(\frac{d^{n-1}u_{j}}{dt^{n-1}}*k^{'}_{j})=g(t)$$

these are algebrical euqations. after you solve them you must integrate and then you
get n functions:
$$k_{j}(t)=\int(k^{'}_{j})dt+c_{j}$$

and you put them back in the "linear"-combination

$$u_{general}(t)=\sum^{n}_{j=1}(k_j(t)*u_{j}(t))$$

8. Oct 25, 2009

### Clef

okay. I think i've sort of got a handle on this way.
Would it be

=A'cos4t +B'sin4t = 0
-A'4sin4t + B'cos4t = -292sin4t

Then by equating coefficients and integrating
=A' = 73
therefore A = -73t
B' = 0
therefore B= 0

therefore the PS is -73tsin4t????

Last edited: Oct 25, 2009