Single-integral solution to 2nd order inhomogeneous ODE

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    2nd order Ode
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Homework Statement



I want to show that $$f''(x) = g(x)$$ has a solution of the form $$f(x) = 2\int_0^{x} dx' (x-x') g(x').$$ It's not hard to verify that it is a solution, the question is how to find it. This should be easy and is likely a standard problem but I haven't found the right approach.

The Attempt at a Solution



Direct integration obviously gives an expression with two nested integrals. I've tried using an integrating factor, writing $$f(x) = x h(x),$$ but that also gives me an expression with two nested integrals. The above diff eq becomes
$$
x h''(x) + h'(x) = g(x)
$$
and so
$$
(x h'(x))' = g(x)
$$
which can be integrated directly, giving ##h'(x)##, which must be integrated again to obtain an expression for ##f##:

$$
f(x) = x \int_0^{x} \frac{dx'}{x'} \int_0^{x'} x'' g(x'') dx''.
$$
Using a more general integrating factor, ##f = \phi h## with ##\phi''=0##, does not help. I'm not sure how to obtain an equation for ##f## with just a single integral. Any suggestions appreciated.
 
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What you are looking for is a derivation of the Green’s function of ##d^2/dx^2##.
 
Also, if you have checked that the expression is a solution, what stops you from following the differentiation steps in reverse?

Your approach with the integrating factor really does not help at all. You are just managing to put the equation on a form that you then put back on the original form. Integrating factors are useful when you have several terms that you want to rewrite as the derivative of a single term and ##f’’## is already a single term.
 
blendecho said:

Homework Statement



I want to show that $$f''(x) = g(x)$$ has a solution of the form $$f(x) = 2\int_0^{x} dx' (x-x') g(x').$ It's not hard to verify that it is a solution, the question is how to find it. This should be easy and is likely a standard problem but I haven't found the right approach.

The Attempt at a Solution



Direct integration obviously gives an expression with two nested integrals. I've tried using an integrating factor, writing $$f(x) = x h(x),$$ but that also gives me an expression with two nested integrals
...

Your formula for ##f(x)## should not have a "2" in it; the correct formula is $$f(x) = \int_0^{x} dx' (x-x') g(x').$$
If you really DO take the second derivative, you will see that this is the case. Furthermore, it can be obtained right away from the nested integral, just by changing the order of integration appropriately.
 
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Orodruin: did you mean the Poisson equation? The point of my integrating factor was to reduce the order of the diff eq, of course the LHS was already a total derivative.

Ray: changing the order of integration on the original nested integral (from naively integrating twice) did the trick and makes the generalization to ##f^{(n)}(x)=g(x)## obvious. Yeesh, it's been a long time since mvar. Thanks for pointing out the factor of 2, too.
 
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