Chapter 7.1, Problem 4E

a.

To determine

To find : The value of $k$ for which the function $y=\cos kt$ satisfy the given differential equation.

Answer to Problem 4E

  $k=\pm \left(\frac{5}{2}\right)$

Explanation of Solution

Given information :

  $y=\cos kt\mathrm{..........}\left(i\right)$

  $y^{″}=-25y\mathrm{.........}\left(ii\right)$

Formula used :

  $\frac{d}{dt}\left(\cos kt\right)=-k\sin kt$ ; $\frac{d}{dt}\left(\sin kt\right)=k\cos kt$

Calculation :

  $y=\cos kt$

Differentiating successively two times with respect to $t$ we get,

  $\begin{array}{l}{y}^{\prime }=-k\sin kt\\ y^{″}=-k^2\cos kt\end{array}$

  $\begin{array}{l}\text{Now, }4y^{″}=-25y\\ \Rightarrow -4k^2\cos kt=-25\cos kt\\ \Rightarrow k^2=\frac{25}{4}\\ \Rightarrow k=\pm \left(\frac{5}{2}\right)\end{array}$

b.

To determine

To show : Every member of the family of the functions $y=A\sin kt+B\cos kt$ is also a solution of the given differential equation $2y^{″}+y^{\prime }-y=0$ .

Explanation of Solution

Given information :

  $2y^{″}+y^{\prime }-y=0$

  $y=A\sin kt+B\cos kt$

Formula used :

  $\frac{d}{dt}\left(\cos kt\right)=-k\sin kt$ ; $\frac{d}{dt}\left(\sin kt\right)=k\cos kt$

Calculation :

When, $y=\frac{5}{2}$

  $y=A\sin \left(\frac{5t}{2}\right)+B\cos \left(\frac{5t}{2}\right)$

  $\begin{array}{l}\Rightarrow y^{\prime }=\frac{dy}{dt}=\left(\frac{5}{2}\right)\left[A\cos \left(\frac{5t}{2}\right)-B\sin \left(\frac{5t}{2}\right)\right]\\ \Rightarrow y^{″}=\frac{d^{2}y}{d{x}^2}={\left(\frac{5}{2}\right)}^2\left[-A\sin \left(\frac{5t}{2}\right)-B\cos \left(\frac{5t}{2}\right)\right]\end{array}$

  $\begin{array}{l}\therefore 4y^{″}=\frac{-25}{\cancel{4}}\times \cancel{4}\left[A\sin \left(\frac{5t}{2}\right)+B\cos \left(\frac{5t}{2}\right)\right]\\ 4y^{″}=-25y\text{ (Verified)}\end{array}$

When, $k=-\frac{5}{2}$

  $y=A\sin \left(-\frac{5t}{2}\right)+B\cos \left(-\frac{5t}{2}\right)$

  $\begin{array}{l}y=A\sin \left(-\frac{5t}{2}\right)+B\cos \left(-\frac{5t}{2}\right)\\ \Rightarrow y=\left[-A\sin \left(\frac{5t}{2}\right)+B\cos \left(\frac{5t}{2}\right)\right]\\ \Rightarrow y^{\prime }=\left(\frac{5}{2}\right)\left[-A\cos \left(\frac{5t}{2}\right)-B\sin \left(\frac{5t}{2}\right)\right]\\ \Rightarrow y^{″}={\left(\frac{5}{2}\right)}^2\left[A\sin \left(\frac{5t}{2}\right)-B\cos \left(\frac{5t}{2}\right)\right]\end{array}$

  $\begin{array}{l}\therefore 4y^{″}=-5^2\left[-A\sin \left(\frac{5t}{2}\right)+B\cos \left(\frac{5t}{2}\right)\right]\\ 4y^{″}=-25y\text{ (Verified)}\end{array}$