Question

In Problems 35 and 36 determine whether it is possible to find values ${\mathit{y}}_{\mathbf{\circ }}$and ${\mathit{y}}_{\mathbf{1}}$(Problem 35) and values of$\mathit{L}\mathbf{}\mathbf{>}\mathbf{}\mathbf{0}$ (Problem 36 ) so that the given boundary-value problem has (a) precisely one nontrivial solution, (b) more than one solution, (c) no solution, (d) the trivial solution.

35. $\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{16}\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}{\mathit{y}}_{\mathbf{\circ }}\mathbf{,}\mathit{y}\mathbf{(}\mathbf{\pi }\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{=}{\mathbf{y}}_{\mathbf{1}}$

Short answer

(a) Doesn't have unique solution.

(b) Infinitely many solutions.

(c) No Solutions.

(d) Solution will not be unique.

Step-by-step solution

To Find the solution of the differential equation

(a) The general solution of the differential equation is $y=\left\{c\_1\right\}\cos 4x+\left\{c\_2\right\}\sin 4x$.

From role="math" localid="1663917273559" $y_{\circ }=y\left(0\right)=c1$we see that $y=\left\{y\_0\right\}\cos 4x+\left\{c\_2\right\}\sin 4x.$

From $y_1=y(\mathrm{\pi }/2)={\mathrm{y}}_0$we see that any solution must satisfy $y_0=y_1$. We also see that when $y_0=y_1$and $y=y_{0}cos4x+c_{2}sin4x$is a solution of the boundary-value problem for any choice of $c_2.$Thus the boundary-value problem does not have a unique solution for any choice of $y_{\circ }$and $y_1$.

Find the solution

(b) Whenever $y_0=y_1$there are infinitely many solutions.

Find the solution

(c) When $y_0{y}_1$ there will be no solutions.

Find the solution

(d) The boundary-value problem will have the trivial solution when

$y_0=y_1=0,$

This solution will not be unique.

Final proof

(a) Doesn't have unique solution.

(b) Infinitely many solutions.

(c) No Solutions.

(d) Solution will not be unique.