Chapter 4.RP, Problem 1RP

In Problems 1-28, find a general solution to the given differential equation.

$y^{\prime \prime }+8y^{\prime }-9y=0$

To determine

To Find:

A general solution to the given differential equation.

Answer to Problem 1RP

Solution:

A general solution to the given differential equation is $c_1{e}^t+c_2{e}^{-9t}$ .

Explanation of Solution

Given:

$y^{\prime \prime }+8y^{\prime }-9y=0$

Approach:

The differential equation

$y^{\prime \prime }+8y^{\prime }-9y=0$ (1)

is homogeneous.

The characteristic equation to (1) is given by

$a{r}^2+br+c=0$

$a$ is a coefficient of $y^{\prime \prime }$,

$b$ is a coefficient of $y^{\prime }$ and,

$c$ is a coefficient of $y$.

Linearly independent solutions of (1) are

$\begin{array}{rll}{y}_1\left(t\right)& =& e^{r_{1}t}\\ y_2\left(t\right)& =& e^{r_{2}t} \left(2\right)\end{array}$

The general solution to (1) is given by

$y\left(t\right)=c_1{y}_1\left(t\right)+c_2{y}_2\left(t\right)$ (3)

$c_1$ and $c_2$ are constants.

Calculation:

From (1) substitute 1 for $a$, 8 for $b$ and $-9$ for $c$ in the characteristic equation,

$\begin{array}{rll}a{r}^2+br+c& =& 0\\ r^2+8r-9& =& 0\\ r_1& =& 1\\ r_2& =& -9 \left(4\right)\end{array}$

From (4) substitute $1$ for $r_1$ and $-9$ for $r_2$ in (2)

$\begin{array}{rll}{y}_1\left(t\right)& =& e^t\\ y_2\left(t\right)& =& e^{-9t}\end{array}$

Substitute $y_1\left(t\right)$ and $y_2\left(t\right)$ in (3)

$\begin{array}{rll}y\left(t\right)& =& c_1{y}_1\left(t\right)+c_2{y}_2\left(t\right)\\ & =& c_1{e}^t+c_2{e}^{-9t}\end{array}$

Conclusion:

Hence, a general solution to the given differential equation is $c_1{e}^t+c_2{e}^{-9t}$.