Mathematical Methods for Physics and Engineering - Matematica.NET

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSy(0) = y(π/2) = 0 is given byy(x) =∫ π/20= − cos xG(x, z)coseczdz∫ x0sin z cosec zdz− sin x= −x cos x +sinx ln(sin x),∫ π/2xcos z cosec zdzwhich agrees with the result obtained in the previous subsections. ◭As mentioned earlier, once a Green’s function has been obtained for a givenLHS and boundary conditions, it can be used to find a general solution for anyRHS; thus, the solution of d 2 y/dx 2 + y = f(x), with y(0) = y(π/2) = 0, is givenimmediately byy(x) =∫ π/20= − cos xG(x, z)f(z) dz∫ x0sin zf(z) dz − sin x∫ π/2xcos zf(z) dz. (15.68)As an example, the reader may wish to verify that if f(x) =sin2x then (15.68)gives y(x) =(− sin 2x)/3, a solution easily verified by direct substitution. Ingeneral, analytic integration of (15.68) for arbitrary f(x) will prove intractable;then the integrals must be evaluated numerically.Another important point is that although the Green’s function method abovehas provided a general solution, it is also useful for finding a particular integralif the complementary function is known. This is easily seen since in (15.68) theconstant integration limits 0 and π/2 lead merely to constant values by whichthe factors sin x and cos x are multiplied; thus the complementary function isreconstructed. The rest of the general solution, i.e. the particular integral, comesto − ∫ x ,andsodropping the constant integration limits, we can find just the particular integral.For example, a particular integral of d 2 y/dx 2 + y = f(x) that satisfies the aboveboundary conditions is given byfrom the variable integration limit x. Therefore by changing ∫ π/2xy p (x) =− cos x∫ xsin zf(z) dz +sinx∫ xcos zf(z) dz.A very important point to realise about the Green’s function method is that aparticular G(x, z) applies to a given LHS of an ODE and the imposed boundaryconditions, i.e. the same equation with different boundary conditions will have adifferent Green’s function. To illustrate this point, let us consider again the ODEsolved in (15.68), but with different boundary conditions.514