# MST210/Unit 1/Book A/Homogeneous Second-Order Differential Equations

To find the particular solution to a first-order differential equation, we needed a single condition to solve the initial value problem. For second-order differential equations, we need two conditions. It is sometimes possible to solve second-order differential equations analytically, if we are given an equation of the form $\frac{d^2y}{dx^2} = f(x)$ but often beyond that it is not possible to use analytical solutions.

# Linear Constant-Coefficient Differential Equations

• A second-order differential equation is linear if it can be expressed in the form

$a(x) \frac{d^2y}{dx^2} + b(x) \frac {dy}{dx} + c(x)y = f(x)$

where $$a(x)$$, $$b(x)$$ and $$c(x)$$ are given continuous functions.

• A linear second-order differential equation is constant-coefficient if the functions $$a(x)$$, $$b(x)$$ and $$c(x)$$ are all constant. Therefore, the equation is of the form

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

where $$a \ne 0$$.

• A linear constant-coefficient differential equation is homogeneous if $$\forall x. f(x) = 0$$, and inhomogeneous otherwise.

Exercise 16
1. This is linear and constant-coeffecient.
2. This is linear and constant-coefficient.
3. This is linear and constant-coefficient.
4. This is linear, but not constant-coefficient.
5. This is not linear.
6. This is linear and constant-coefficient.
7. This is linear and constant-coefficient.
8. This is linear and constant-coefficient.
1. This is inhomogeneous.
2. This is inhomogeneous.
3. This is homogeneous.
4. This is inhomogeneous.
5. This is inhomogeneous.
6. This is homogeneous.
1. The dependent variable is the variable we are differentiating with respect to the independent variable. For equations (i) to (v), $$y$$ is dependent and $$x$$ is independent. For (vi) $$t$$ is dependent and $$\theta$$ is independent. In (vii) $$x$$ is dependent and $$t$$ is independent. Finally, in (viii) $$x$$ is dependent and $$t$$ is independent.

# Principle of Superposition

If $$y_1(x)$$ is a solution of the linear second-order differential equation

$a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = f_1(x)$

and $$y_2(x)$$ is a solution of the linear second-order differential equation

$a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = f_2(x)$

then the function

$y(x) = k_1 y_1(x) + k_2 y_2(x)$

where $$k_1$$ and $$k_2$$ are constants, is a solution of the differential equation

$a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = k_1 y_1(x) + k_2 y_2(x)$

# Method of Solution

## The Auxiliary Equation

The auxiliary equation of the homogeneous linear constant-coefficient second-order differential equation

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

is the quadratic equation

$a \lambda ^2 + b \lambda + c = 0$

This arises by considering $$y = Ae^{\lambda x}$$ as a solution, and making the substituitions $$\frac{dy}{dx} = \lambda A e^{\lambda x}$$ and $$\frac{d^y}{dx^2} = \lambda ^2 A e ^{\lambda x}$$:

$\begin{array}{lcl} a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy &=& a (\lambda ^2 A e^{\lambda x}) + b (\lambda A e ^{\lambda x}) + c (Ae ^ {\lambda x}) \\ &=& (a \lambda^2 + b\lambda + c) Ae^{\lambda x}\\ &=& 0 \end{array}$

This shows $$y = Ae^{\lambda x}$$ is a solution, provided the aforemention quadratic equation holds (as $$A$$ is an arbitrary constant).

Exercise 17
1. $$\lambda^2 - 5 \lambda + 6 = 0$$
2. $$\lambda^2 - 9 = 0$$
3. $$\lambda^2 + 2 \lambda = 0$$

### Two Distinct Real Solutions

As the auxiliary equation is a quadratic equation, there are a few options for possible solutions. In the case of two real solutions, $$\lambda_1$$ and $$\lambda_2$$, these roots correspond to $$y = Ce^{\lambda_1 x}$$ and $$y = De^{\lambda_2 x}$$, where $$C$$ and $$D$$ are arbitrary constants.

Example 8
1. The auxiliary equation is $\lambda^2 - 3 \lambda + 2 = 0$.

2. We can rewrite the auxiliary equation as

$(\lambda - 1)(\lambda - 2) =$

Thus we have $$\lambda = 1$$ and $$\lambda = 2$$.

Therefore the two solutions to the equation are $$y = Ce^x$$ and $$y = De^{2x}$$

1. We note that $\frac{d^2y}{dx^2} - 3 \frac{dy}{dx} + 2y = 0$ is a linear second order differential-equation, with two solutions $y_1(x) = Ce^x$ and $y_2(x) = De^{2x}$

Therefore, by the Principle of Superposition, $y = k_1 Ce^{x} + De^{2x}$ is also a solution.

However, in the book, we show that this is a solution by directly substituting in and evaluating.

It can be seen that for if $$\lambda_1$$ and $$\lambda_2$$ are distinct roots of the auxiliary equation of a homogeneous linear constant-coefficient second-order differential equation, then any solution of the form

$y = Ce^{\lambda_1 x} + De^{\lambda_2x}$

for some choice of constants $$C$$ and $$D$$, is also a solution.

Exercise 18
1. The auxiliary equation is

$\lambda^2 + 5\lambda + 6 = 0$.

We can rewrite this as $$(\lambda + 2)(\lambda + 3) = 0$$, and so there are two solutions: $$\lambda_1 = -2$$ and $$\lambda_2 = -3$$.

Therefore the general solution is given by

$y = Ce^{-2x} + De^{-3x}$

Where $$C$$ and $$D$$ are arbitrary constants.

2. The auxiliary equation is

$2\lambda^2 + 3\lambda = 0$

That is, $$\lambda(2\lambda + 3) = 0$$, so there are two solutions: $$\lambda_1 = 0$$ and $$\lambda_2 = -\frac{3}{2}$$.

Therefore the general solution is given by $y = C + De^{-\frac{3x}{2}}$

where $$C$$ and $$D$$ are arbitrary constants.

3. The auxiliary equation is

$\lambda^2 - 4 = 0$

Therefore there are two distinct solutions: $$\lambda_{1,2} = \pm 2$$.

Thus the general solution is

$y = Ce^{2u} + De^{-2u}$

where $$C$$ and $$D$$ are arbitrary constants

### Equal Real Roots

In general, when the two solutions of the auxiliary equation $$\lambda_{1,2}$$ are equal ($$\lambda_1 = \lambda_2$$), there there are general solutions of the form

$y = Ce^{\lambda_1 x} + Dxe^{\lambda_1 x} = (C + Dx)e^{\lambda_1 x}$

where $$C$$ and $$D$$ are arbitrary constants.

Exercise 19
1. The auxiliary equation is

$\lambda^2 + 2\lambda + 1 = 0$

This can be factored into $$(\lambda + 1)^2$$, so we have two equal solutions $$\lambda_{1,2} = -1$$.

Therefore, the general solution is

$y = (C + Dx)e^{-x}$

where $$C$$ and $$D$$ are arbitrary constants.

2. The auxiliary equation is

$\lambda^2 - 4 \lambda + 4 = 0$

This equation can be factored as $$(\lambda - 2)^2$$, and so we have two equal solutions $$\lambda_{1,2} = 2$$.

Therefore, the general solution is

$s = (C + Dt)e^{2t}$ where $$C$$ and $$D$$ are arbitrary constants.

### Complex-Valued Solutions

The final possibility for solutions of the auxiliary equation is a pair of complex conjugate roots, $$\lambda_1 = \alpha + \beta i$$, $$\lambda_2 = \alpha - \beta i$$. In this case, we can use the same form for the solution where $$\lambda_{1,2}$$ are distinct real solutions, but if we allow the constants to be complex, we can arrive at a real solution via Euler’s formula.

In this case, the general solution (following this process) is:

$y = e^{\alpha x} (C \cos \beta x + D \sin \beta x)$

where $$C$$ and $$D$$ are arbitrary constants.

Exercise 20
1. The auxiliary equation is $\lambda^2 + 4 \lambda + 8 = 0$

By solving this quadratic equation by the quadratic formula, we have

$\begin{array}{lcl} \lambda &=& \frac{-4 \pm \sqrt{4^2 - 32}}{2} \\ &=& \frac{-4 \pm 4i}{2} \\ &=& -2 \pm 2i \end{array}$

Thus the auxiliary equation has a pair of complex conjugate roots. Therefore the general solution is

$y = e^{-2x}(C \cos 2x + D \sin 2x)$ where $$C$$ and $$D$$ are arbitrary constants.

2. The auxiliary equation is $\lambda^2 + 9 = 0$

Thus we have $$\lambda_{1,2} = \pm 3i$$. These solutions are a pair of complex conjugate roots, therefore the general solution is

$\theta = C \cos 3t + D \sin 3t$ where $$C$$ and $$D$$ are arbitrary constants.

Exercise 21
1. The auxiliary equation is $\lambda^2 + 4 = 0$ Thus the solutions are a pair of complex conjugate roots, $$\lambda_{1,2} = \pm 2i$$. Following Procedure 5, the general solution is

$y_1 = C \cos 2 x + D \sin 2 x$ where $$C$$ and $$D$$ are arbitrary constants.

2. The auxiliary equation is $\lambda^2 - 6 \lambda + 8 = 0$ This can be factored as $$(\lambda - 4)(\lambda - 2) = 0$$. Therefore we have two distinct real solutions, $$\lambda_1 = 4$$ and $$\lambda_2 = 2$$. Following Procedure 5, the general solution is

$u = Ce^{4x} + De^{2x}$ where $$C$$ and $$D$$ are arbitrary constants.

3. The auxiliary equation is $\lambda^2 + 2\lambda = 0$ Therefore we have two distinct real solutions, $$\lambda_1 = -2$$ and $$\lambda_2 = 0$$. Following Procedure 5, the general solution is

$y = Ce^{-2x} + D$ where $$C$$ and $$D$$ are arbitrary constants.

4. The auxiliary equation is $\lambda^2 - 2\lambda + 1 = 0$ This can be factored as $(\lambda - 1)^2 = 0$ and so we have a pair of equal real solutions, $$\lambda_{1,2} = 1$$. Therefore, following Procedure 5, the general solution is $y = (C + Dx)e^x$ where $$C$$ and $$D$$ are arbitrary constants.

5. The auxiliary equation is $\lambda^2 - \omega^2 = 0$ This is a difference of two squares, so we have $$(\lambda + \omega)(\lambda - \omega) = 0$$, and so the auxiliary equation has two distinct real solutions, $$\lambda_{1,2} = \pm \omega$$. Following Procedure 5, the general solution is $y = Ce^{\omega x} + De^{-\omega x}$ where $$C$$ and $$D$$ are arbitrary constants.

6. The auxiliary equation is $\lambda^2 + 4 \lambda + 29 = 0$

By the quadratic formulua, we have

$\begin{array}{lcl} \lambda &=& \frac{-4 \pm \sqrt{4^2 - 4 \cdot 29}}{2} \\ &=& \frac{-4 \pm \sqrt{-100}}{2} \\ &=& -2 \pm 5i \end{array}$

Following Procedure 5, the general solution is $y = e^{-2x} (C \cos 5 x + D \sin 5 x)$

Exercise 22

We can rearrange the differential equation into a homogeneous differential equation, $\ddot{\theta} + \frac{g}{l}\theta = 0$

The auxiliary equation for this differential equation is $\lambda^2 + \frac{g}{l} = 0$

Therefore, the solution to this eqution is a pair of complex conjugate roots $$\lambda_{1,2} = \pm \sqrt{\frac{g}{l}} i$$.

Following Procedure 5, the general solution is therefore

$\theta = C \cos (\sqrt{\frac{g}{l}} t) + D \sin (\sqrt{\frac{g}{l}} t)$