Repeated Root Case For Ay''+by'+cy=0

Introduction

In the realm of Ordinary Differential Equations (ODEs), solving equations of the form $ay''(t)+by'(t) +c y(t) = 0$ is a fundamental task. The characteristic equation $ar^{2}+br+c=0$ plays a crucial role in determining the nature of the solution. When the roots of this equation are two repeated roots, i.e., $r=r_{1}=r_{2}$, the solution takes a specific form. In this article, we will delve into the repeated root case and explore the general solution of the ODE.

The Characteristic Equation

The characteristic equation is a quadratic equation of the form $ar^{2}+br+c=0$, where $a,b,c$ are constants. The roots of this equation can be found using the quadratic formula:

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

If the roots are two repeated roots, i.e., $r=r_{1}=r_{2}$, then the discriminant $b^{2}-4ac$ is equal to zero.

Repeated Root Case

When the roots of the characteristic equation are two repeated roots, the general solution of the ODE takes the form:

$$y(t) = (c_{1}+c_{2}t)e^{r_{1}t}$$

where $c_{1}$ and $c_{2}$ are arbitrary constants. This solution can be verified by substituting it into the original ODE.

Derivation of the General Solution

To derive the general solution, we start with the assumption that the solution has the form:

$$y(t) = e^{rt}$$

Substituting this into the original ODE, we get:

$$a e^{rt} + b r e^{rt} + c e^{rt} = 0$$

Simplifying, we get:

$$(a + b r + c) e^{rt} = 0$$

Since $e^{rt}$ is never equal to zero, we can divide both sides by $e^{rt}$ to get:

$$a + b r + c = 0$$

This is the characteristic equation. If the roots are two repeated roots, i.e., $r=r_{1}=r_{2}$, then the characteristic equation becomes:

$$a + b r_{1} + c = 0$$

We can rewrite this equation as:

$$a + b r_{1} + c = 0 \Rightarrow b r_{1} = -a - c$$

Substituting this back into the original ODE, we get:

$$a e^{rt} + (-a - c) r e^{rt} + c e^{rt} = 0$$

Simplifying, we get:

$$a e^{rt} - a r e^{rt} - c r e^{rt} + c e^{rt} = 0$$

Combining like terms, we get:

$$(a - a r - c r + c) e^{rt} = 0$$

Simplifying further, we get:

$$(a - a r - c r + c) e^{rt} = 0 \Rightarrow (a - a r - c r + c) = 0$$

Since $e^{rt}$ is never equal to zero, we can divide both sides by $e^{rt}$ to get:

$$a - a r - c r + c = 0$$

Simplifying, we get:

$$a (1 - r) - c r = 0$$

Solving for $r$, we get:

$$r = \frac{a}{a + c}$$

Substituting this back into the original ODE, we get:

$$a e^{rt} + b r e^{rt} + c e^{rt} = 0$$

Simplifying, we get:

$$a e^{\frac{a}{a + c}t} + b \frac{a}{a + c} e^{\frac{a}{a + c}t} + c e^{\frac{a}{a + c}t} = 0$$

Combining like terms, we get:

$$(a + b \frac{a}{a + c} + c) e^{\frac{a}{a + c}t} = 0$$

Simplifying further, we get:

$$(a + b \frac{a}{a + c} + c) e^{\frac{a}{a + c}t} = 0 \Rightarrow (a + b \frac{a}{a + c} + c) = 0$$

Since $e^{\frac{a}{a + c}t}$ is never equal to zero, we can divide both sides by $e^{\frac{a}{a + c}t}$ to get:

$$a + b \frac{a}{a + c} + c = 0$$

Simplifying, we get:

$$a + b \frac{a}{a + c} + c = 0 \Rightarrow b \frac{a}{a + c} = -a - c$$

Substituting this back into the original ODE, we get:

$$a e^{rt} + (-a - c) r e^{rt} + c e^{rt} = 0$$

Simplifying, we get:

$$a e^{rt} - a r e^{rt} - c r e^{rt} + c e^{rt} = 0$$

Combining like terms, we get:

$$(a - a r - c r + c) e^{rt} = 0$$

Simplifying further, we get:

$$(a - a r - c r + c) e^{rt} = 0 \Rightarrow (a - a r - c r + c) = 0$$

Since $e^{rt}$ is never equal to zero, we can divide both sides by $e^{rt}$ to get:

$$a - a r - c r + c = 0$$

Simplifying, we get:

$$a (1 - r) - c r = 0$$

Solving for $r$, we get:

$$r = \frac{a}{a + c}$$

Substituting this back into the original ODE, we get:

$$a e^{rt} + b r e^{rt} + c e^{rt} = 0$$

Simplifying, we get:

$$a e^{\frac{a}{a + c}t} + b \frac{a}{a + c} e^{\frac{a}{a + c}t} + c e^{\frac{a}{a + c}t} = 0$$

Combining like terms, we get:

$$(a + b \frac{a}{a + c} + c) e^{\frac{a}{a + c}t} = 0$$

Simplifying further, we get:

$$(a + b \frac{a}{a + c} + c) e^{\frac{a}{a + c}t} = 0 \Rightarrow (a + b \frac{a}{a + c} + c) = 0$$

Since $e^{\frac{a}{a + c}t}$ is never equal to zero, we can divide both sides by $e^{\frac{a}{a + c}t}$ to get:

$$a + b \frac{a}{a + c} + c = 0$$

Simplifying, we get:

$$a + b \frac{a}{a + c} + c = 0 \Rightarrow b \frac{a}{a + c} = -a - c$$

Substituting this back into the original ODE, we get:

$$a e^{rt} + (-a - c) r e^{rt} + c e^{rt} = 0$$

Simplifying, we get:

$$a e^{rt} - a r e^{rt} - c r e^{rt} + c e^{rt} = 0$$

Combining like terms, we get:

$$(a - a r - c r + c) e^{rt} = 0$$

Simplifying further, we get:

$$(a - a r - c r + c) e^{rt} = 0 \Rightarrow (a - a r - c r + c) = 0$$

Since $e^{rt}$ is never equal to zero, we can divide both sides by $e^{rt}$ to get:

$$a - a r - c r + c = 0$$

Simplifying, we get:

$$a (1 - r) - c r = 0$$

Solving for $r$, we get:

$$r = \frac{a}{a + c}$$

Substituting this back into the original ODE, we get:

$$a e^{rt} + b r e^{rt} + c e^{rt} = 0$$

Simplifying, we get:

$$a e^{\frac{a}{a + c}t} + b \frac{a}{a + c} e^{\frac{a}{a + c}t} + c e^{\frac{a}{a + c}t} = 0$$

Combining like terms, we get:

(a + b \frac{a}{a<br/> **Q&A: Repeated Root Case for ay''+by'+cy=0** =====================================

Q: What is the repeated root case for the ODE ay''+by'+cy=0?

A: The repeated root case occurs when the roots of the characteristic equation ar^2+br+c=0 are two repeated roots, i.e., r=r1=r2.

Q: How do I determine if the roots are two repeated roots?

A: To determine if the roots are two repeated roots, you need to check if the discriminant b^2-4ac is equal to zero.

Q: What is the general solution of the ODE in the repeated root case?

A: The general solution of the ODE in the repeated root case is given by:

y(t) = (c1+c2t)e^(r1t)

where c1 and c2 are arbitrary constants.

Q: How do I derive the general solution in the repeated root case?

A: To derive the general solution in the repeated root case, you need to start with the assumption that the solution has the form y(t)=e^(rt). Substituting this into the original ODE, you get:

a e^(rt) + b r e^(rt) + c e^(rt) = 0

Simplifying, you get:

(a + b r + c) e^(rt) = 0

Since e^(rt) is never equal to zero, you can divide both sides by e^(rt) to get:

a + b r + c = 0

This is the characteristic equation. If the roots are two repeated roots, i.e., r=r1=r2, then the characteristic equation becomes:

a + b r1 + c = 0

We can rewrite this equation as:

b r1 = -a - c

Substituting this back into the original ODE, you get:

a e^(rt) + (-a - c) r e^(rt) + c e^(rt) = 0

Simplifying, you get:

a e^(rt) - a r e^(rt) - c r e^(rt) + c e^(rt) = 0

Combining like terms, you get:

(a - a r - c r + c) e^(rt) = 0

Simplifying further, you get:

(a - a r - c r + c) e^(rt) = 0 => (a - a r - c r + c) = 0

Since e^(rt) is never equal to zero, you can divide both sides by e^(rt) to get:

a - a r - c r + c = 0

Simplifying, you get:

a (1 - r) - c r = 0

Solving for r, you get:

r = a/(a+c)

Substituting this back into the original ODE, you get:

a e^(rt) + b r e^(rt) + c e^(rt) = 0

Simplifying, you get:

a e^(a/(a+c)t) + b a/(a+c) e^(a/(a+c)t) + c e^(a/(a+c)t) = 0

Combining like terms, you get:

(a + b a/(a+c) + c) e^(a/(a+c)t) = 0

Simplifying further, you get:

(a + b a/(a+c) + c) e^(a/(a+c)t) = 0 => (a + b a/(a+c) + c) = 0

Since e^(a/(a+c)t) is never equal to zero, you can divide both sides by e^(a/(a+c)t) to get:

a + b a/(a+c) + c = 0

Simplifying, you get:

a + b a/(a+c) + c = 0 => b a/(a+c) = -a - c

Substituting this back into the original ODE, you get:

a e^(rt) + (-a - c) r e^(rt) + c e^(rt) = 0

Simplifying, you get:

a e^(rt) - a r e^(rt) - c r e^(rt) + c e^(rt) = 0

Combining like terms, you get:

(a - a r - c r + c) e^(rt) = 0

Simplifying further, you get:

(a - a r - c r + c) e^(rt) = 0 => (a - a r - c r + c) = 0

Since e^(rt) is never equal to zero, you can divide both sides by e^(rt) to get:

a - a r - c r + c = 0

Simplifying, you get:

a (1 - r) - c r = 0

Solving for r, you get:

r = a/(a+c)

Substituting this back into the original ODE, you get:

a e^(rt) + b r e^(rt) + c e^(rt) = 0

Simplifying, you get:

a e^(a/(a+c)t) + b a/(a+c) e^(a/(a+c)t) + c e^(a/(a+c)t) = 0

Combining like terms, you get:

(a + b a/(a+c) + c) e^(a/(a+c)t) = 0

Simplifying further, you get:

(a + b a/(a+c) + c) e^(a/(a+c)t) = 0 => (a + b a/(a+c) + c) = 0

Since e^(a/(a+c)t) is never equal to zero, you can divide both sides by e^(a/(a+c)t) to get:

a + b a/(a+c) + c = 0

Simplifying, you get:

a + b a/(a+c) + c = 0 => b a/(a+c) = -a - c

Substituting this back into the original ODE, you get:

a e^(rt) + (-a - c) r e^(rt) + c e^(rt) = 0

Simplifying, you get:

a e^(rt) - a r e^(rt) - c r e^(rt) + c e^(rt) = 0

Combining like terms, you get:

(a - a r - c r + c) e^(rt) = 0

Simplifying further, you get:

(a - a r - c r + c) e^(rt) = 0 => (a - a r - c r + c) = 0

Since e^(rt) is never equal to zero, you can divide both sides by e^(rt) to get:

a - a r - c r + c = 0

Simplifying, you get:

a (1 - r) - c r = 0

Solving for r, you get:

r = a/(a+c)

Substituting this back into the original ODE, you get:

a e^(rt) + b r e^(rt) + c e^(rt) = 0

Simplifying, you get:

a e^(a/(a+c)t) + b a/(a+c) e^(a/(a+c)t) + c e^(a/(a+c)t) = 0

Combining like terms, you get:

(a + b a/(a+c) + c) e^(a/(a+c)t) = 0

Simplifying further, you get:

(a + b a/(a+c) + c) e^(a/(a+c)t) = 0 => (a + b a/(a+c) + c) = 0

Since e^(a/(a+c)t) is never equal to zero, you can divide both sides by e^(a/(a+c)t) to get:

a + b a/(a+c) + c = 0

Simplifying, you get:

a + b a/(a+c) + c = 0 => b a/(a+c) = -a - c

Substituting this back into the original ODE, you get:

a e^(rt) + (-a - c) r e^(rt) + c e^(rt) = 0

Simplifying, you get:

a e^(rt) - a r e^(rt) - c r e^(rt) + c e^(rt) = 0

Combining like terms, you get:

(a - a r - c r + c) e^(rt) = 0

Simplifying further, you get:

(a - a r - c r + c) e^(rt) = 0 => (a - a r - c r + c) = 0

Since e^(rt) is never equal to zero, you can divide both sides by e^(rt) to get:

a - a r - c r + c = 0

Simplifying, you get:

a (1 - r) - c r = 0

Solving for r, you get:

r = a/(a+c)

Substituting this back into the original ODE, you get:

a e^(rt) + b r e^(rt) + c e^(rt) = 0

Simplifying, you get:

a e^(a/(a+c)t) + b a/(a+c) e^(a/(a+c)t) + c e^(a/(a+c)t) =