Question

For what values of \(r\) does the function \(y = {e^{rx}}\) satisfy the differential equation \(y'' - 4y' + y = 0\).

Short answer

The value of \(r\) is \(2 \pm \sqrt 3 \).

Step-by-step solution

The Chain Rule

The composite function \(F = f \circ g\), or \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiableat \(x\) and \(F'\) is obtained as:

\(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\)

In Leibniz notation, the derivative is shown below:

\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}\)

Determine the value of \(r\)

Use the chain rule to differentiate the function \(y\) as shown below:

\(\begin{aligned}y' &= \frac{d}{{dx}}{e^{rx}}\\ &= {e^{rx}}\frac{d}{{dx}}\left( {rx} \right)\\ &= r{e^{rx}}\end{aligned}\)

Use the chain rule to differentiate the function \(y'\) as shown below:

\(\begin{aligned}y'' &= \frac{d}{{dx}}\left( {r{e^{rx}}} \right)\\ &= r{e^{rx}}\frac{d}{{dx}}\left( {rx} \right)\\ &= r{e^{rx}} \cdot r\\ &= {r^2}{e^{rx}}\end{aligned}\)

Use the expression of \(y,y',\) and \(y''\) in the differential equation \(y'' - 4y' + y = 0\) to obtain as shown below:

\(\begin{aligned}y'' - 4y' + y &= {r^2}{e^{rx}} - 4r{e^{rx}} + {e^{rx}}\\ &= {e^{rx}}\left( {{r^2} - 4r + 1} \right)\end{aligned}\)

Since \({e^{rx}} \ne 0\) using the quadratic formula to obtain the value of \(r\) as shown below:

\(\begin{aligned}r &= \frac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}}\\ &= \frac{{4 \pm \sqrt {16 - 4} }}{2}\\ &= \frac{{4 \pm \sqrt {12} }}{2}\\ &= 2 \pm \sqrt 3 \end{aligned}\)

Thus, the value of \(r\) is \(2 \pm \sqrt 3 \).