Solution for a second-order differential equation

chaksome
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Homework Statement
Second order ODE
Relevant Equations
x‘’(t)+4x’(t)+8x(t) = f(t)
I wish to know if there is a method to work out x(t).
[No matter which form f(t) is]
Thank you~
 
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You know PF: what's your attempt ? Know about Laplace transforms ?
 
BvU said:
You know PF: what's your attempt ? Know about Laplace transforms ?
Oh, I haven‘t learned that^v^
Do you mean this chapter will help me understand this problem?
I can try to learn it~
 
It gives you a formal, general method to find a solution, yes.
 
First, the "associated homogenous equation" is x''+ 8x'+ 4x= 0 which has "characteristic equation" r^2+ 8r+ 4= 0. Solve that by "completing the square": r^2+ 8r+ 16- 16+ 4= (r+ 4)^2- 12= 0 so (r+ 4)^2= 12, r= -4\pm2\sqrt{3}. The general solution to the "associate homogeneous equation" is e^{-4x}\left(Ae^{2x\sqrt{3}}+ Be^{-2x\sqrt{3}}\right).

Now use "variation of parameters". We look for a solution to the entire equation of the form y(x)= e^{-4x}\left(u(x)e^{2x\sqrt{3}}+ v(x)e^{-2x\sqrt{3}}\right). That is, we let the "parameters", A and B, vary. Surely That method is given in your textbook. You can also read about it at http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx.

The calculations are tedious but doable! The result will be an integral in f(x) that you add to the general solution to the homogeneous equation.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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