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 The Solution to the Differential Equation u"(x) − u(x) = f(x)

The differential equation

#### uxx - u(x) = f(x)

is a linear inhomogeneous ordinary differential equation. Its solution can be obtained most easily using Fourier or Laplace Transforms.

Taking the Laplace Transform of the above equation gives:

#### (s²−1)U = F

where U is the Laplace transform of u(x) and F is the Laplace transform of f(x) and where it is assumed that u(0)=0 and u'(0)=0.

This means that

#### U = (s²−1)-1F = −½[1/(S+1) - 1/(S-1)]F

The Laplace transform of sinh(x) is equal to −½[1/(S+1) - 1/(S-1)] so

#### U = Lsinh(x)F.

By the convolution theorem then

#### u(x) = ∫0xsinh(x−z)f(z)dz

To verify that this a solution, first

#### u'(x) = sinh(0)f(x) + ∫0xcosh(x−z)f(z)dz and since sinh(0)=0 this reduces to u'(x) = ∫0xcosh(x−z)f(z)dz

The second derivative is given by

#### u"(x) = cosh(0)f(x) + ∫0xsinh(x−z)f(z)dzand since cosh(0)=1 u"(x) = f(x) + ∫0xsinh(x−z)f(z)dzand since and therefore u"(x)−u(x) = f(x).

The lower limit of 0 for the integral is arbitrary, stemming from the way the Laplace transform has to be defined. It is appropriate to take the lower limit as −∞ and to take u(−∞)=0 and u'(−∞)=0. Thus the solution is