Second Order Differential Equations

Second-order differential equations

A second-order differential equation is an equation in which the highest derivative of the dependent variable with respect to the independent variable is of order two. The general second-order ordinary differential equation has the form:

$p(x, y)\frac{d^2y}{dx^2} + q(x, y)\frac{dy}{dx} + r(x, y)y = g(x, y)$

A linear second-order differential equation takes the form:

$p(x)\frac{d^2y}{dx^2} + q(x)\frac{dy}{dx} + r(x)y = g(x) \quad (7.12)$

where $p(x)$, $q(x)$, $r(x)$, and $g(x)$ are continuous functions of $x$ or constants in a given interval, and $p(x)$ is not identically zero.

If $g(x) = 0$, the differential equation is called homogeneous. If $g(x) \ne 0$, it is called non-homogeneous.

Example 1

Classify each of the following differential equations as homogeneous or non-homogeneous:

(a) $\frac{d^2y}{dx^2} - y = 0$

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

(c) $xy'' + xy' = 3$

(d) $y'' + 5y' - 4y = 5x$

(e) $x^2y''' + y'' - e^xy' = 1$

(f) $\frac{d^2y}{dx^2} - \sin x \frac{dy}{dx} + \cos x \, y = 3$

Solution:

(a) Homogeneous ($g(x) = 0$)

(b) Non-homogeneous ($g(x) = 2\sin x \ne 0$)

(c) Non-homogeneous ($g(x) = 3 \ne 0$)

(d) Non-homogeneous ($g(x) = 5x \ne 0$)

(e) Non-homogeneous ($g(x) = 1 \ne 0$)

(f) Non-homogeneous ($g(x) = 3 \ne 0$)

Characteristic equation of a second-order homogeneous differential equation

Consider a second-order homogeneous differential equation with constant coefficients:

$a_2y'' + a_1y' + a_0y = 0 \quad (7.13)$

where $a_2$, $a_1$, and $a_0$ are constants.

Assume a solution of the form $y = e^{\lambda x}$, where $\lambda$ is a constant.

Then, $y' = \lambda e^{\lambda x}$ and $y'' = \lambda^2 e^{\lambda x}$.

Substituting these into equation (7.13) gives:

$a_2\lambda^2 e^{\lambda x} + a_1\lambda e^{\lambda x} + a_0e^{\lambda x} = 0$

$e^{\lambda x}(a_2\lambda^2 + a_1\lambda + a_0) = 0$

Since $e^{\lambda x} \ne 0$, we have:

$a_2\lambda^2 + a_1\lambda + a_0 = 0 \quad (7.14)$

Equation (7.14) is called the characteristic equation or auxiliary equation.

Example 2

Write the characteristic equation for each of the following differential equations:

(a) $y'' + 6y' - 8y = 0$

(b) $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0$

(c) $y'' - 9y = 0$

(d) $2\frac{d^2y}{dx^2} + \frac{dy}{dx} - y = 0$

Solution:

(a) $\lambda^2 + 6\lambda - 8 = 0$

(b) $\lambda^2 - 2\lambda + 1 = 0$

(c) $\lambda^2 - 9 = 0$

(d) $2\lambda^2 + \lambda - 1 = 0$

Roots of characteristic equations and solutions to homogeneous differential equations

Consider the second-order homogeneous differential equation $a_2y'' + a_1y' + a_0y = 0$, whose characteristic equation is $a_2\lambda^2 + a_1\lambda + a_0 = 0$. The roots of the characteristic equation determine the solution of the differential equation. Using the quadratic formula, the roots are:

$\lambda = \frac{-a_1 \pm \sqrt{a_1^2 - 4a_2a_0}}{2a_2}$

Three cases are considered:

Case 1: Distinct real roots

If $a_1^2 - 4a_2a_0 > 0$, there are two distinct real roots, $\lambda_1$ and $\lambda_2$. The linearly independent solutions are $y_1 = e^{\lambda_1 x}$ and $y_2 = e^{\lambda_2 x}$. The general solution is:

$y = Ae^{\lambda_1 x} + Be^{\lambda_2 x}$

where $A$ and $B$ are arbitrary constants.

Case 2: Repeated real roots

If $a_1^2 - 4a_2a_0 = 0$, there are two equal real roots, $\lambda_1 = \lambda_2 = \lambda$. The solutions $y_1 = e^{\lambda x}$ and $y_2 = e^{\lambda x}$ are not linearly independent.

A second linearly independent solution is found using reduction of order. Let $y_2 = xe^{\lambda x}$. The general solution is:

$y = Ae^{\lambda x} + Bxe^{\lambda x} = e^{\lambda x}(A + Bx)$

Case 3: Conjugate complex roots

If $a_1^2 - 4a_2a_0 < 0$, there are two conjugate complex roots, $\lambda_1 = \alpha + i\beta$ and $\lambda_2 = \alpha - i\beta$, where $i = \sqrt{-1}$.

The linearly independent solutions are $e^{(\alpha + i\beta)x}$ and $e^{(\alpha - i\beta)x}$. Using Euler's formula ($e^{i\theta} = \cos\theta + i\sin\theta$), we can rewrite the general solution as:

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

where $A$ and $B$ are arbitrary constants.

Example 1

Solve each of the following differential equations:

(a) $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - 15y = 0$

(b) $y'' - 4y' + 4y = 0$; $y(0) = 4$ and $y'(0) = 10$

(c) $\frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 25y = 0$

Solution:

(a) The characteristic equation is $\lambda^2 + 2\lambda - 15 = 0$. Factoring gives $(\lambda + 5)(\lambda - 3) = 0$. The roots are $\lambda_1 = 3$ and $\lambda_2 = -5$. The general solution is:

$y = c_1e^{3x} + c_2e^{-5x}$

(b) The characteristic equation is $\lambda^2 - 4\lambda + 4 = 0$, or $(\lambda - 2)^2 = 0$. The repeated root is $\lambda = 2$. The general solution is:

$y = e^{2x}(A + Bx)$

Using the initial conditions:

$y(0) = 4$: $4 = e^0(A + B(0)) \Rightarrow A = 4$

$y'(x) = 2e^{2x}(A + Bx) + Be^{2x}$

$y'(0) = 10$: $10 = 2(4) + B \Rightarrow B = 2$

The particular solution is $y = e^{2x}(4 + 2x)$.

(c) The characteristic equation is $\lambda^2 - 6\lambda + 25 = 0$. Using the quadratic formula:

$\lambda = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(25)}}{2} = \frac{6 \pm \sqrt{36 - 100}}{2} = \frac{6 \pm \sqrt{-64}}{2} = 3 \pm 4i$

The roots are $\lambda = 3 \pm 4i$. The general solution is:

$y = e^{3x}(A\cos(4x) + B\sin(4x))$

Example 2

Solve the boundary-value problem $\frac{d^2x}{d\theta^2} + 4\frac{dx}{d\theta} + 29x = 0$; $x(0) = 4$ and $x'(\pi/2) = -7e^{-\pi}$.

Solution:

The characteristic equation is $\lambda^2 + 4\lambda + 29 = 0$. Using the quadratic formula:

$\lambda = \frac{-4 \pm \sqrt{4^2 - 4(1)(29)}}{2} = \frac{-4 \pm \sqrt{16 - 116}}{2} = \frac{-4 \pm \sqrt{-100}}{2} = -2 \pm 5i$

The general solution is:

$x(\theta) = e^{-2\theta}(A\cos(5\theta) + B\sin(5\theta))$

Using the boundary conditions:

$x(0) = 4$: $4 = e^0(A\cos(0) + B\sin(0)) \Rightarrow A = 4$

$x'(\theta) = -2e^{-2\theta}(A\cos(5\theta) + B\sin(5\theta)) + e^{-2\theta}(-5A\sin(5\theta) + 5B\cos(5\theta))$

$x'(\pi/2) = -7e^{-\pi}$:

$-7e^{-\pi} = -2e^{-\pi}(4\cos(5\pi/2) + B\sin(5\pi/2)) + e^{-\pi}(-5(4)\sin(5\pi/2) + 5B\cos(5\pi/2))$

$-7e^{-\pi} = -2e^{-\pi}(0+B) + e^{-\pi}(-20+0)$

$-7 = -2B - 20$

$13 = -2B$

$B = -\frac{13}{2}$

The particular solution is:

$x(\theta) = e^{-2\theta}\left(4\cos(5\theta) - \frac{13}{2}\sin(5\theta)\right)$