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.

$\mathbf{36}\mathbf{.}{\mathit{y}}^{\mathbf{\varphi }\mathbf{}\mathbf{\varphi }}\mathbf{+}\mathbf{16}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\left(}\mathit{L}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

(a) Will be a unique solution when $\sin 4\ne 0;$ that is, when $L\ne k\pi /4$ where $k=1,2,3,......$

(b) There will be infinitely many solutions when$\sin 4L=0$and that is, when where .

(c) There will be no solution when $sin4L\ne 0$and $1-cos4L\ne 0;$that is, when $L=k\pi /4$ where $k=1,3,5,\mathrm{.....}$.

(d) There can be no trivial solution since it would fail to satisfy the boundary conditions.

Step-by-step solution

To Find the solution of the differential equation

(a) The general solution of the differential equation is $y=c_{1}cos4x+c_{2}sin4x$. From $y_0=y\left(0\right)=c1$ we see that $y=y_{0}cos4x+c_{2}sin4x.$From $1=y\left(L\right)=cos4L+c_{2}sin4L$ we see that $c_2=(1-cos4L)/sin4L.$ Thus

$y\left(x\right)=cos4x+\frac{1-cos4L}{sin4L}sin4x$ will be a unique solution when $sin4\ne 0;$ that is, when $L\ne k\pi /4$ where $k=1,2,3,.....$

Find the solution

(b) There will be infinitely many solutions when $\sin 4L=0$ and $1-\cos 4L=0;$ that is, when $L=k\pi /2$ where $k=1,2,3,....$

Find the solution

(c) There will be no solution when $sin4L\ne 0$ and $1-cos4L\ne 0;$ that is, when $L=k\pi /4$ where $k=1,3,5,.....$

Find the solution

(d) There can be no trivial solution since it would fail to satisfy the boundary conditions.

Final proof

(a) Will be a unique solution when $\sin 4\ne 0;$ that is, when $L\ne k\pi /4$ where $k=1,2,3,......$

(b) There will be infinitely many solutions when $\sin 4L=0$and $1-\cos 4L=0;$ that is, when $L=k\pi /2$ where $k=1,2,3,....$.

(c) There will be no solution when $sin4L\ne 0$and $1-cos4L\ne 0;$ that is, when $L=k\pi /4$ where $k=1,3,5,.....$

(d) There can be no trivial solution since it would fail to satisfy the boundary conditions.