Chapter 17.1, Problem 19E

To determine

To solve:

The initial value problem ${9y}^{\text{'}\text{'}}+12y^{\text{'}}+4y=0 , y\left(0\right)=1, y^{\text{'}}\left(0\right)=0$

Solution:

$y=e^{-\frac{2}{3}x}+\frac{2}{3}x{e}^{-\frac{2}{3}x}$

Explanation:

1) Concept:

i) Theorem: If $y_1$ and $y_2$ are linearly independent solutions of the second order Homogeneous linear differential equation $a{y}^{\text{'}\text{'}}+b{y}^{\text{'}}+cy=0$ on an interval, and $a\ne 0$, then the general solution is given by $y\left(x\right)=c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)$ where $c_1 and c_2$ are constants.

ii) In general solution of $a{y}^{\text{'}\text{'}}+b{y}^{\text{'}}+cy=0$

Roots of $a{r}^2+br+c=0$	General solution
$r_1, r_2$ are real and distinct	$y=c_1{e}^{r_{1}x}+c_2{e}^{r_{2}x}$
$r_1=r_2=r$	$y=c_1{e}^{rx}+c_{2}x{e}^{rx}$
$r_1 , r_2 complex :\alpha \pm i\beta$	$y=e^{\alpha x}(c_{1}cos\beta x+c_{2}sin\beta x$)

2) Given:

The initial value problem ${9y}^{\text{'}\text{'}}+12y^{\text{'}}+4y=0 , y\left(0\right)=1, y^{\text{'}}\left(0\right)=0$

3) Calculations:

The auxiliary equation of the differential equation ${9y}^{\text{'}\text{'}}+12y^{\text{'}}+4y=0$ is $9r^2+12r+4=0$

Solving the auxiliary equation  $9r^2+12r+4=0$

$9r^2+12r+4=0$

$9r^2+6r+6r+4=0$

$3r(3r+2)+2(3r+2)=0$

$(3r+2)(3r+2)=0$

$r=\frac{-2}{3},\frac{-2}{3}$

By given concept general solution of differential equation ${9y}^{\text{'}\text{'}}+12y^{\text{'}}+4y=0$ is

$y=c_1{e}^{-\frac{2}{3}x}+c_{2}x{e}^{-\frac{2}{3}x}$

Differentiate this equation with respect to $x$

$y=-{\frac{2}{3}c}_1{e}^{-\frac{2}{3}x}+c_2{e}^{-\frac{2}{3}x}{-\frac{2}{3}c}_{2}x{e}^{-\frac{2}{3}x}$

Applying initial conditions $y\left(0\right)=1, y^{\text{'}}\left(0\right)=0$

$y\left(0\right)=c_1$ Therefore $, c_1=1$

$y^{\text{'}}\left(0\right)=-\frac{2}{3}{c}_1+c_2$ Therefore,  $-\frac{2}{3}{c}_1+c_2=0$ this imply $c_2=\frac{2}{3}$

Thus, the solution of the initial value problem $y^{\text{'}\text{'}}-2y^{\text{'}}-3y=0 , y\left(0\right)=1, y^{\text{'}}\left(0\right)=0$ is $y=e^{-\frac{2}{3}x}+\frac{2}{3}x{e}^{-\frac{2}{3}x}$

Conclusion:

The solution of the initial value problem $y^{\text{'}\text{'}}-2y^{\text{'}}-3y=0 , y\left(0\right)=2, y^{\text{'}}\left(0\right)=2$ is $y=e^{-\frac{2}{3}x}+\frac{2}{3}x{e}^{-\frac{2}{3}x}$