Q1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Show that if y(t) satisfies $\mathbf{y}\mathbf{"}\mathbf{-}\mathbf{ty}\mathbf{=}\mathbf{0}$, then y(-t) satisfies $\mathbf{y}\mathbf{"}\mathbf{+}\mathbf{ty}\mathbf{=}\mathbf{0}$.

Step-by-Step Solution

Answer

Thus, it is proved that if  $\mathrm{y}\left(\mathrm{t}\right)$ satisfies $\mathrm{y}"-\mathrm{ty}=0$, then $\mathrm{y}(-\mathrm{t})$ satisfies $\mathrm{y}"+\mathrm{ty}=0.$

Step 1: General form

Chain rule of the derivative states that:

$\frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\right]=\mathrm{f}\text{'}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\times \mathrm{g}\text{'}\left(\mathrm{x}\right)$

Step 2: Evaluate the given equation.

Given that $\mathrm{y}\left(\mathrm{t}\right)$ satisfies $\mathrm{y}"-\mathrm{ty}=0$.

 

To prove that  $\mathrm{y}(-\mathrm{t})$ satisfies $\mathrm{y}"+\mathrm{ty}=0$.

 

Let us assume that, $\mathrm{u}\left(\mathrm{t}\right)=\mathrm{y}(-\mathrm{t})$.

 

Using the chain rule find the first and second derivatives of $\mathrm{u}\left(\mathrm{t}\right)$.

Case (1):

$\begin{array}{c}\mathrm{u}\text{'}\left(\mathrm{t}\right)=\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{y}(-\mathrm{t})\right]\\ =\mathrm{y}\text{'}(-\mathrm{t})\times (-\mathrm{t})\text{'}\\ =-\mathrm{y}\text{'}(-\mathrm{t})\end{array}$

Case (2):

$\begin{array}{c}\mathrm{u}"\left(\mathrm{t}\right)=[-\mathrm{y}\text{'}(-\mathrm{t})]\text{'}\\ =-\mathrm{y}"(-\mathrm{t})\times (-\mathrm{t})\text{'}\\ =\mathrm{y}"(-\mathrm{t})\end{array}$

Step 3: Substitute the values.

Substitute $\mathrm{t}$  within $\mathrm{y}"+\mathrm{ty}=0.$

$\begin{array}{c}\mathrm{y}"(-\mathrm{t})-(-\mathrm{t})\mathrm{y}(-\mathrm{t})=0\\ \mathrm{y}"(-\mathrm{t})+\mathrm{ty}(-\mathrm{t})=0\end{array}$

Substitute the result of cases (1) and (2);

 

$\mathrm{u}"\left(\mathrm{t}\right)+\mathrm{tu}\left(\mathrm{t}\right)=0$

 

Therefore,$\mathrm{u}\left(\mathrm{t}\right)=\mathrm{y}(-\mathrm{t})$ satisfies the equation $\mathrm{y}"+\mathrm{ty}=0$.