SecondOrder Nonhomogenous Linear Differential Equations

[wubie]

Hello,

First here is the question that I am supposed to solve:

Solve the following nonhomogeneous differential equations:

b) y'' + 2y' + 2y = e^-x

c) 2y'' + y' = cos 2x.


I am supposed to be using the method of variation of parameters to solve these equations. What my problem is I end up getting to a point where I have two equations in which I should be able to solve for the derivative of parameter one (u'sub1) and the derivative of parameter two (u'sub2). Unfortunately I am getting stuck. And I am not sure why.

For b) I have the following two equations:

u'sub1 ysub1 + u'sub2 ysub2 = 0 = u'sub1 e^-x + u'sub2 xe^-x

and the particular equation

u'sub2 - u'sub2 x + usub2 x - u'sub1 + usub1 = 1

From these I am supposed to find u'sub1 and u'sub2 and eventually come to find usub1 and usub2.

Now when I solve for

u'sub1 e^-x + u'sub2 xe^-x = 0

I get u'sub1 = (-u'sub2 xe^-x)/e^-x = -u'sub2 x

I then sub. into the other equation for u'sub1

u'sub2 - u'sub2 x + usub2 x - u'sub1 + usub1 = 1

becomes

u'sub2 - u'sub2 x + usub2 x + u'sub2 x + usub1 = 1

which becomes

u'sub2 + usub2 x + usub1 = 1

But I am stumped here. How do I solve for u'sub2 when I still have usub2 and usub1? I know I am missing something incredibly obvious. I just can't seem to know what.

For question b) I am having similar problems - still trying to solve for u'sub1 and u'sub2.


Update: I have figured out question c). I am still working on part b however.

2nd Update: I figured out question b as well. Thanks to all who took the time to look at my post.

Cheers.

 

[anathema]

Thank you for your question. It seems like you are on the right track with your approach using the method of variation of parameters. However, there are a few errors in your calculations that are causing you to get stuck.

For part b), the first equation you have is correct: u'sub1 ysub1 + u'sub2 ysub2 = 0 = u'sub1 e^-x + u'sub2 xe^-x. However, when you solve for u'sub1, you should get u'sub1 = (u'sub2 xe^-x)/e^-x = u'sub2 x. The negative sign in your solution is incorrect.

Next, when you substitute this value for u'sub1 into the second equation, you should get u'sub2 + u'sub2 x + u'sub1 = 1. This is where your mistake lies - you have mistakenly added an extra u'sub2 x term. The correct equation is u'sub2 + u'sub1 = 1. Now, you can solve for u'sub2 and get u'sub2 = 1 - u'sub1.

For part c), it seems like you have already solved it. Great job! Just a small note - when you substitute the value for u'sub1 into the second equation, you should get u'sub2 + u'sub1 = 1 instead of u'sub2 + u'sub2 = 1.

I hope this helps you in solving part b) as well. Keep up the good work and don't hesitate to ask for help if you get stuck again. Good luck!

 

[M98Ranger]

Hello,

Glad to hear that you were able to figure out both parts of the question. The method of variation of parameters can be tricky, so it's understandable that you got stuck at first. Here are some tips that might help you in the future:

1. When solving for u'sub1 and u'sub2, make sure to isolate them on one side of the equation and all other terms on the other side. This will make it easier to solve for them individually.

2. When substituting u'sub1 into the other equation, make sure to substitute it as a whole, not just part of it. In your example, you only substituted u'sub1 as "u'sub2 x", but you should have substituted it as "-u'sub2 x". This will result in a different equation and may help you solve for u'sub2.

3. Remember to use the initial conditions given in the question to solve for the constants u'sub1 and u'sub2. These conditions will help you determine the specific values of the parameters and complete the solution.

I hope these tips will help you in your future problem-solving. Keep practicing and you will become more comfortable with the method of variation of parameters. Good luck!