I'm self studying DE's and found some notes on the Internet.(adsbygoogle = window.adsbygoogle || []).push({});

If I understood well, S-L problem is to find the solution to [itex]Ly=(py')'-qy=f[/itex] where y is a function of [itex]x[/itex].

Getting to the point and skipping tons of details, if [itex]u_1 (x)[/itex] and [itex]u_2 (x)[/itex] are linearly independent solutions to the homogeneous S-L problem, i.e. [itex]Ly=0[/itex], then a solution to [itex]Ly=f[/itex] is given by ... [itex]y(x)=\int _a^b g(x,t)f(t)dt[/itex] where [itex]g(x,t)[/itex] is the green function associated to the operator L, namely [itex]g(x,t)=\frac{1}{p(a)W(a)}[/itex] multiplied by [itex]u_1 (x) u_2 (t)[/itex] if a [itex]\leq x \leq t[/itex] or [itex]u_1 (t) u_2 (x)[/itex] if [itex]t <x<b[/itex].

Where [itex]W(a)[/itex] is the Wronksian of [itex]u_1 (a)[/itex] and [itex]u_2 (a)[/itex].

So it seems that in order to solve the original S-L DE, I must first solve the corresponding homogeneous DE. Then, calculate the Wronskian of these 2 L.I. solutions. Then form [itex]g(x,t)[/itex] and finally perform the integral that has no reason to be easy to solve.

My question is... is this method (via Green function+Wronskian) really simpler than just using variation of coefficient method, once I solved the homogeneous S-L DE?

In other words, once I get [itex]u_1 (x)[/itex] and [itex]u_2(x)[/itex], is seeking a solution to the non homogeneous DE of the form [itex]y(x)=u_1 (x) v_1(x)+u_2(x)v_2 (x)[/itex] really harder than using Green function method?!

P.S.:I attach the document I studied on.

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