General solution to inhomogeneous second order equation

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SUMMARY

The solution to the inhomogeneous second-order differential equation x'' + cx' = f(t) involves first solving the homogeneous part, yielding A + B*exp(-ct). The particular solution is derived using the integral (1/c)int((1-exp(c(s-t))f(s))ds from 0 to t. An alternative method involves transforming the equation using Fourier Transforms, leading to an algebraic equation for \tilde(x)(t) and subsequently applying the inverse Fourier Transform. Additionally, by letting v = x', the equation simplifies to a linear form v' + cv = f(t), which can be solved using an integrating factor e^{ct}.

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  • Understanding of second-order differential equations
  • Familiarity with integrating factors in differential equations
  • Knowledge of Fourier Transforms and their applications
  • Basic calculus, particularly integration techniques
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Students and professionals in mathematics, physics, and engineering who are solving differential equations, particularly those dealing with inhomogeneous second-order equations.

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Homework Statement

I need to find the solution to x'' + cx' = f(t), for a general f.

Homework Equations


The Attempt at a Solution



Obviously first I solve the homogeneous part to give me A + B*exp(-ct). I also know that the particular solution is written as (1/c)int((1-exp(c(s-t))f(s))ds where the integral is between 0 and t. However I am not sure why this is so, any help would be much appreciated.
 
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Have you learned about Fourier Transforms yet? If so, just transform both sides of the DE, solve the resulting algebraic equation for \tilde(x)(t) and then take the inverse Fourier Transform.
 
Oh, dear! Using "Fourer Series" for this is like using a shotgun to kill a fly!

Let v= x' and your differential equation becomes v'+ cv= f(t). That's a linear equation with "integrating factor" e^{ct}. That is,
\frac{d(e^{ct}v)}{dx}= e^{ct}v'+ ce^{ct}v= e^{ct}f(t)

Integrating both sides,
e^{ct}v= \int_{t_0}^t e^{cs}f(x)ds+ C

From that,
v= x'= Ce^{-ct}+ e^{-ct}\int_{t_0}^t e^{cs}f(s)ds

Now, integrate again:
x(t)= C_1 e^{-ct}+ \int_{t_0}^t\left(e^{-cu}\int_{t_0}^u e^{cs}f(s)ds\right)du+ C_2
 

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