Mathematical Methods for Physics and Engineering - Matematica.NET

23.8 EXERCISESthus Hermitian. In order to solve this inhomogeneous equation using SH theory, however,we must first find the eigenvalues and eigenfunctions of the corresponding homogeneousequation.In fact, we have considered the solution of the corresponding homogeneous equation(23.13) already, in subsection 23.4.1, where we found that it has two eigenvalues λ 1 =2/πand λ 2 = −2/π, with eigenfunctions given by (23.16). The normalised eigenfunctions arey 1 (x) = √ 1 (sin x +cosx) and y 2 (x) = √ 1 (sin x − cos x) (23.58)π πand are easily shown to obey the orthonormality condition (23.49).Using (23.54), the solution to the inhomogeneous equation (23.57) has the formy(x) =a 1 y 1 (x)+a 2 y 2 (x), (23.59)where the coefficients a 1 and a 2 are given by (23.53) with f(x) =sin(x + α). Therefore,using (23.58),1a 1 =1 − πλ/21a 2 =1+πλ/2∫ π0∫ π01√ π(sin z +cosz) sin(z + α) dz =1√ π(sin z − cos z) sin(z + α) dz =√ π(cos α +sinα),2 − πλ√ π(cos α − sin α).2+πλSubstituting these expressions for a 1 and a 2 into (23.59) and simplifying, we find thatthe solution to (23.57) is given byy(x) =11 − (πλ/2) 2 [sin(x + α)+(πλ/2) cos(x − α)]. ◭23.1 Solve the integral equation∫ ∞023.8 Exercisescos(xv)y(v) dv =exp(−x 2 /2)for the function y = y(x) forx>0. Note that for x