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

## Question:

Solve the differential equation {eq}{y}''- 10{y}' + 25y = 0 {/eq} with the initial conditions: when{eq}x = 0, y = 1 {/eq} and when {eq}x = 0, {y}' = 1 {/eq}.

A) Find the general solution.

B) Find the particular solution using the given initial conditions.

C) Sketch the graph. Use the windows {eq}[-2, 1], \, [-2, 2] {/eq}.

## Initial value problem:

The initial value problem is the differential equation with some initial conditions. The differential equation is the equation which contains one or more functions and their derivatives.

## Answer and Explanation: 1

**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}.

#### Learn more about this topic:

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Chapter 11 / Lesson 13Learn to define the initial value problem and initial value formula. Learn how to solve initial value problems in calculus. See examples of initial value problems.