Question

In Problems 39–44, $\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{x}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{x}$is a two-parameter family of solutions of the second-order DE $\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}$. If possible, find a solution of the differential equation that satisfies the given side conditions. The conditions specified at two different points are called boundary conditions.

role="math" localid="1663830656127" $\mathit{y}\mathbf{\text{'}}\mathbf{(}\mathit{\pi }\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathit{\pi }\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

No solution

Step-by-step solution

First boundary condition

Differentiating the family of solutions given us the first derivative,

$y\text{'}=-2c_1\sin 2x+2c_2\cos 2x$

Substituting the first boundary condition to the above derivative of two parameter family of the solution then gives,

$$\begin{gathered} y\text{'}\left(\frac{\pi }{2}\right)=-2c_1\sin \left(2·\frac{\pi }{2}\right)+2c_2\cos \left(2·\frac{\pi }{2}\right) \\ 1=-2c_1\sin \pi +2c_2\cos \pi \\ 1=-2c_1\left(0\right)+2c_2(-1) \end{gathered}$$

Simplify,

$$\begin{gathered} 1=-2c_2 \\ -\frac{1}{2}=c_2 \end{gathered}$$

Second boundary condition

Substituting the secondary boundary condition to the given two parameter family of solution gives,

$$\begin{gathered} y\text{'}\left(\pi \right)=-2c_1\sin \left(2\right(\pi \left)\right)+2c_2\cos \left(2\right(\pi \left)\right) \\ 0=-2c_1\sin \left(2\pi \right)+2c_2\cos \left(2\pi \right) \\ 0=-2c_1\left(0\right)+2c_2\left(1\right) \\ 0=2c_2 \\ 0=c_2 \end{gathered}$$

We see that the constant$c_2$ has two different values for different boundary condition which is not possible.

Hence, there is no solution of the differential equation that satisfies the given border conditions.