Chapter 17, Problem 1RCC

(a) Write the general form of a second-order homogeneous linear differential equation with constant coefficients.

(b) Write the auxiliary equation.

(c) How do you use the roots of the auxiliary equation to solve the differential equation? Write the form of the solution for each of the three cases that can occur.

To determine

To write: The general form of a second-order homogeneous linear differential equation with constant coefficients.

Answer to Problem 1RCC

The general form of a second-order homogeneous linear differential equation with constant coefficients is $\underset{\_}{a\frac{d^{2}y}{d{x}^2}+b\frac{dy}{dx}+cy=0}$.

Explanation of Solution

Formula used:

Consider the second-order linear differential equation as follows.

$P\left(x\right)\frac{d^{2}y}{d{x}^2}+Q\left(x\right)\frac{dy}{dx}+R\left(x\right)y=G\left(x\right)$ (1)

Here,

$P$, $Q$, $R$, and $G$ are continuous functions.

If $G\left(x\right)=0$ for all values of $x$ then second-order differential equation is known as homogeneous equation.

Consider the value of $G\left(x\right)$ as follows.

$G\left(x\right)=0$

Substitute $0$ for $G\left(x\right)$ in equation (1),

$P\left(x\right)\frac{d^{2}y}{d{x}^2}+Q\left(x\right)\frac{dy}{dx}+R\left(x\right)y=0$ (2)

Consider $P\left(x\right)$, $Q\left(x\right)$ and $R\left(x\right)$ has a constant values represented as $a$, $b$ and $c$ respectively.

Substitute $a$ for $P\left(x\right)$, $b$ for $Q\left(x\right)$ and $c$ for $R\left(x\right)$ in equation (2),

$a\frac{d^{2}y}{d{x}^2}+b\frac{dy}{dx}+cy=0$ (3)

Thus, the general form of a second-order homogeneous linear differential equation with constant coefficients is $\underset{\_}{a\frac{d^{2}y}{d{x}^2}+b\frac{dy}{dx}+cy=0}$.

To determine

To write: The auxiliary equation.

Answer to Problem 1RCC

The auxiliary equation of the second-order differential equation is $\underset{\_}{a{r}^2+br+c=0}$.

Explanation of Solution

Modify equation (3) as follows.

$a{y}^{″}+b{y}^{\prime }+cy=0$ (4)

In equation (4), function $y$ such that a constant times its second derivative $\left(y^{″}\right)$ plus second constant times its second derivative $\left(y^{\prime }\right)$ plus a third constant times $y$ is equal to $0$.

Consider a exponential function for $y$ as follows.

$y=e^{rx}$

Differentiate $y$ with respect to $x$.

$y^{\prime }=r{e}^{rx}$

Differentiate $y^{\prime }$ with respect to $x$.

$y^{″}=r^2{e}^{rx}$

Substitute $r^2{e}^{rx}$ for $y^{″}$, $r{e}^{rx}$ for $y^{\prime }$ and $e^{rx}$ for $y$ in equation (4),

$a{r}^2{e}^{rx}+rb{e}^{rx}+c{e}^{rx}=0$

$e^{rx}\left(a{r}^2+br+c\right)=0$ (5)

Since, the value of $e^{rx}$ is never $0$, $e^{rx}$ becomes the solution for $y$ with $r$ is a root of the equation.

$a{r}^2+br+c=0$ (6)

Equation (6) is known as characteristic equation or auxiliary equation of the second-order differential equation $a{y}^{″}+b{y}^{\prime }+cy=0$.

Thus, the auxiliary equation of the second-order differential equation is $\underset{\_}{a{r}^2+br+c=0}$.

To determine

To explain: The use of roots of the auxiliary equation to solve the differential equation and write the form of the solution for each of the three cases.

Answer to Problem 1RCC

The use of roots of the auxiliary equation to solve the differential equation is explained.

The form of the solution for each of the three cases is written.

Explanation of Solution

Formula used:

Consider the second-order differential equation as follows.

$a{x}^2+bx+c=0$

Write the expression for quadratic formula.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x$ consists of two roots represented by $x_1$ and $x_2$ as follows.

$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ (7)

$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$ (8)

Modify equation (7) as follows.

$r_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$

Modify equation (8) as follows.

$r_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

$r_1$ and $r_2$ are the roots of auxiliary equation $a{r}^2+br+c=0$.

Thus, the use of roots of the auxiliary equation to solve the differential equation is explained.

Three different cases are obtained depending upon the term $b^2-4ac$ present in the roots of auxiliary equation.

Case I:

Consider the value of $b^2-4ac>0$.

In this case, the roots of auxiliary equation are real and distinct. So two linear independent solutions are occurs such as $y_1=e^{r_{1}x}$ and $y_2=e^{r_{2}x}$.

Write the expression for general solution.

$y=c_1{y}_1+c_2{y}_2$ (9)

Here,

$c_1$ and $c_2$ are constants.

Substitute $e^{r_{1}x}$ for $y_1$ and $e^{r_{2}x}$ for $y_2$,

$y=c_1{e}^{r_{1}x}+c_2{e}^{r_{2}x}$

Case II:

Consider the value of $b^2-4ac=0$.

In this case, the roots of auxiliary equation are real and equal.

Consider $r$ be the common roots for $r_1$ and $r_2$. So the solutions are occurs such as $y_1=e^{rx}$ and $y_2=x{e}^{rx}$.

Substitute $e^{rx}$ for $y_1$ and $x{e}^{rx}$ for $y_2$ in equation (9),

$y=c_1{e}^{rx}+x{c}_2{e}^{rx}$

Case III:

Consider the value of $b^2-4ac<0$.

In this case, the roots of auxiliary equation are complex numbers.

Consider $r_1$ and $r_2$ are represented as follows.

$\begin{array}{l}{r}_1=\alpha +i\beta \\ r_2=\alpha -i\beta \end{array}$

Write the expression for general solution with complex roots.

$y=e^{\alpha x}\left(c_1\cos \beta x+c_2\sin \beta x\right)$

Thus, the form of the solution for each of the three cases is written.