Solve the differential equation y double prime - 10y prime + 25y = 0 with the initial conditions:...

Given:

• The given differential equation is, {eq}y'' - 10y' + 25y = 0 {/eq} and the initial conditions are {eq}y\left( 0 \right) = 1 {/eq} and {eq}y'\left( 0 \right) = 1 {/eq}.

A

First write the characteristic equation of the given differential equation and find its roots.

{eq}\begin{align*} {m^2} - 10m + 25 &= 0\\ {m^2} - 5m - 5m + 25 &= 0\\ \left( {m - 5} \right)\left( {m - 5} \right) &= 0\\ m &= 5,5 \end{align*} {/eq}

Here the roots are real and equal.

If the roots are real and equal, then the general solution will be of the form {eq}y\left( x \right) = {c_1}{e^{mx}} + {c_2}x{e^{mx}} {/eq}.

Therefore, the general solution is {eq}y\left( x \right) = {c_1}{e^{5x}} + {c_2}x{e^{5x}} {/eq}.

B

Substitute the initial condition {eq}y\left( 0 \right) = 1 {/eq} in the general solution.

{eq}\begin{align*} y\left( 0 \right) &= {c_1}{e^{5\left( 0 \right)}} + {c_2}\left( 0 \right){e^{5\left( 0 \right)}}\\ 1 &= {c_1} + 0\\ {c_1} &= 1 \end{align*} {/eq}

Now compute the first derivative of the general solution.

{eq}\begin{align*} y'\left( x \right) &= {c_1}\left( {5{e^{5x}}} \right) + {c_2}\left( {x5{e^{5x}} + {e^{5x}}} \right)\\ &= 5{c_1}{e^{5x}} + {c_2}\left( {5x{e^{5x}} + {e^{5x}}} \right) \end{align*} {/eq}

Substitute the initial condition {eq}y'\left( 0 \right) = 1 {/eq} and {eq}{c_1} = 1 {/eq} in the above equation.

{eq}\begin{align*} y'\left( 0 \right) &= 5\left( 1 \right){e^{5\left( 0 \right)}} + {c_2}\left( {5\left( 0 \right){e^{5\left( 0 \right)}} + {e^{5\left( 0 \right)}}} \right)\\ 1 &= 5 + {c_2}\left( {0 + 1} \right)\\ {c_2} &= 1 - 5\\ {c_2} &= - 4 \end{align*} {/eq}

Thus, the required particular solution is, {eq}y\left( x \right) = {e^{5x}} - 4x{e^{5x}} {/eq}.

C

Use online graphing calculator and sketch the graph of {eq}y\left( x \right) = {e^{5x}} - 4x{e^{5x}} {/eq} in the window {eq}- 2 \le x \le 1 {/eq} and {eq}- 2 \le y \le 2 {/eq}.