Second Order Differential Equation Question

TheSetMiner

Solve the following second order linear dierential equation
x'' + 4x = 2 cos(t);
subject to the following initial conditions
x(0) = 0;
x' (0) = 0:

Does anyone know how to go about this. Previously I would get the characteristic equation as x^2 + ax + b = 0 but I am struggling to see how I can bring the cost term into this because I don't actually know its value and hence won't be able to get the roots of the equation. Furthermore the ODE given doesn't have an x' term which seems a little strange? Any help would be greatly appreciated.

CJC86

TheSetMiner wrote: »

Solve the following second order linear dierential equation
x'' + 4x = 2 cos(t);
subject to the following initial conditions
x(0) = 0;
x' (0) = 0:

Does anyone know how to go about this. Previously I would get the characteristic equation as x^2 + ax + b = 0 but I am struggling to see how I can bring the cost term into this because I don't actually know its value and hence won't be able to get the roots of the equation. Furthermore the ODE given doesn't have an x' term which seems a little strange? Any help would be greatly appreciated.

Use the same method as usual to find the general solution (ie. the solution to the homogeneous equation, x'' + 4x = 0), remembering that if the roots of the quadratic are complex (a+bi, a-bi) then the solution is Ae^(at)cos(bt) + Be^(at)sin(bt).

Then for the particular solution, I think the way I would approach it is to take a good guess at the solution. So, if the RHS is an exponential, let your particular solution = Ae^(at), and if it's trigonometric then let your particular solution = Asin(at) + Bcos(at).

TheSetMiner

solving x'' + 4x = 0 ie the quadratic x^2 = -4 gives x = 2i

am i right in saying that the solution will be something like Acos2t + Bsin2t + Xp then?

I really am a little more than stuck on getting the particular solution. In my notes I have loads of formula and notation but your idea to guess it makes more sense and seems easier.
So in this case because there is trig in the formula, my Xp would be Asin(at) + Bcos(at). So then I have a lot of constants. how do I tie these up from the 2cost term initially in the formula?

thanks for your help!

Michael Collins

You are correct. You should try Xp = Asin(at) + Bcos(at), but you know the value of 'a' in this equation, since it must be the same as the forcing function on the Right-Hand-Side of the differential equation, i.e. you get 'a' from 2cos(t) = 2cos(1t), so 'a' is...?

So now your full solution is

x(t) = Acos2t + Bsin2t + Xp

and when plugged into the differential equation this must equal

cos(t).

But you know, by definition, that the homogeneous solution must become zero when plugged into the differential equation, so all you need to do is make sure

Xp'' + 4Xp = 2cos(t).

So in order words, form the above equation, and choose the constants to make it true.

Then put it all together and you can find the homogeneous constants from the initial conditions.

Michael Collins

Actually, just be careful with your constants. The A's and B's are different constants for the homogeneous solution vs the particular solution - but I think you know this!

TheSetMiner

Ok I can set this equation up and I know a is 1 but then I am left with 4 constants in my homogenous solution? namely two in the general and two in the particular solution. I assume there is a way of getting the constants for my particular before using the initial conditions to get the two for the general.

So how do I use the 2cost term to get the constants for my particular solution?

thanks for replying by the way

CJC86

TheSetMiner wrote: »

Ok I can set this equation up and I know a is 1 but then I am left with 4 constants in my homogenous solution? namely two in the general and two in the particular solution. I assume there is a way of getting the constants for my particular before using the initial conditions to get the two for the general.

So how do I use the 2cost term to get the constants for my particular solution?

thanks for replying by the way

You should have only 2 undetermined constants from the general solution, the particular solution never has undetermined constants, because it is not a homogeneous solution (I hope that makes sense to you).

Anyway, if Xp = Ccost + Dsint is your particular solution, get that by letting Xp''+4Xp=2cost. Then add your general solution and particular solution and set them equal to the initial values to determine the other two constants.

TheSetMiner

but when you say Ccost + Dsint is the particular solution, aren't C and D constants there too? Are they given to us from the 2cost term is it? How do I get C and D from it?

eoins23456

http://onsolver.com/diff-equation.php this might help

Michael Collins

TheSetMiner wrote: »

but when you say Ccost + Dsint is the particular solution, aren't C and D constants there too? Are they given to us from the 2cost term is it? How do I get C and D from it?

Make an equation using the particular solution:

Xp = Ccost + Dsint

So now

Xp'' + 4Xp = 2cos(t).

So differentiate Xp twice, add it to 4 times Xp and this must equal 2cos(t). So equate the coefficients of cos and sin, this will give you C and D.

Blackpanther95

You might need to try a variation :Xp=Atcost+Btsint. On rare occasions the particular solution would be a simple linear comnination of the homogenous solution. In this case you need an unusual paetical solution. As in the case of forced harmonic motion (resonance solution).But thats only if ur unlucky with ur Xh