Solve the boundary value problem [latex]y^{\prime\prime}+y=0,\quad y(0)=1,\quad y(\pi)=a,[/latex] (10) where a is a given number.

The general solution of this differential equation is [latex]y=c_{1}\cos x+c_{2}\sin x,[/latex] (11) and from the first boundary condition we find that [latex]c_{1}[/latex] = 1. The second boundary condition now requires that −[latex]c_{1}[/latex] = a. These two conditions on [latex]c_{1}[/latex] are incompatible if a ≠ −1, so the problem has no solution in that case. However, if a = −1, then both boundary conditions are satisfied provided that [latex]c_{1}[/latex] = 1, regardless of the value of [latex]c_{2}[/latex]. In this case there are infinitely many solutions of the form [latex]y=\cos x+c_{2}\sin x,[/latex] where [latex]c_{2}[/latex] remains arbitrary. This example illustrates that a nonhomogeneous boundary value problem may have no solution---and also that under special circumstances it may have infinitely many solutions.

Question 10.1.2: Solve the boundary value problem y′′ + y = 0, y(0) = 1, y(π)......

Elementary Differential Equations and Boundary Value Problems [1529447]

Solve the boundary value problem

$y^{\prime\prime}+y=0,\quad y(0)=1,\quad y(\pi)=a,$ (10)

where a is a given number.

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The general solution of this differential equation is

$y=c_{1}\cos x+c_{2}\sin x,$ (11)

and from the first boundary condition we find that $c_{1}$ = 1. The second boundary condition now requires that − $c_{1}$ = a. These two conditions on $c_{1}$ are incompatible if a ≠ −1, so the problem has no solution in that case. However, if a = −1, then both boundary conditions are satisfied provided that $c_{1}$ = 1, regardless of the value of $c_{2}$ . In this case there are infinitely many solutions of the form

$y=\cos x+c_{2}\sin x,$

where $c_{2}$ remains arbitrary. This example illustrates that a nonhomogeneous boundary value problem may have no solution—and also that under special circumstances it may have infinitely many solutions.