Introduction To Differential Equations

The order of a differential equation is the highest order derivative present in the equation. The degree of a differential equation is the power of the highest order derivative, after the equation has been made free from radicals and fractions involving the derivatives.

Example 1

(a) $\frac{d^2y}{dx^2} + \frac{dy}{dx} + y = 0$ is a second-order differential equation of degree one.

(b) $(y'')^2 + y' + y^3 = 0$ is a second-order differential equation of degree two.

(c) $\frac{dy}{dx} + xy = 0$ is a first-order differential equation of degree one.

(d) $(y'')^3 + 4(y')^4 - 5y^3 = \cos 2x$ is a second-order differential equation of degree three.

A differential equation is linear if the dependent variable and its derivatives appear only to the first power, and there are no products of the dependent variable and its derivatives, or products of derivatives of different orders. The coefficients can be functions of the independent variable or constants.

An $n$th order linear ordinary differential equation in the dependent variable $y$ and the independent variable $x$ has the form:

$a_n(x)\frac{d^ny}{dx^n} + a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}} + \dots + a_1(x)\frac{dy}{dx} + a_0(x)y = g(x)$

where $a_n$ is not identically zero, and $a_n(x), a_{n-1}(x), \dots, a_0(x), g(x)$ are functions of $x$ or constants.

A linear ordinary differential equation satisfies these conditions:

1. The dependent variable $y$ and all its derivatives occur to the first degree only.
2. No product of the dependent variable $y$ and any of its derivatives is present.
3. No transcendental functions (trigonometric, logarithmic, exponential, etc.) of $y$ occur.

If any of these conditions are not met, the differential equation is non-linear.

Example 2

Examples of linear differential equations:

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

(b) $\ln x \frac{dy}{dx} + e^y = x$. Note: This is non-linear due to the $e^y$ term.

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

(d) $6\frac{d^2y}{dx^2} + 9\frac{dy}{dx} + e^xy = x^2$

Example 3

Examples of non-linear differential equations:

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

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

(c) $xy\left(\frac{dy}{dx}\right)^2 = -7\frac{dy}{dx}$

(d) $x\frac{dy}{dx} + \sin y = 0$

Example 4

In each of the following equations, indicate the independent and dependent variables:

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

(b) $\frac{d^3u}{dy^3} + 3u = 2$

(c) $y' + xy = 0$

(d) $x'' - 2x' + x = t^2 + t + 4$

Solution:

(a) $y$ is the dependent variable, and $x$ is the independent variable.

(b) $u$ is the dependent variable, and $y$ is the independent variable.

(c) $y$ is the dependent variable, and $x$ is the independent variable.

(d) $x$ is the dependent variable, and $t$ is the independent variable.

Example 5

For each of the following differential equations, state its order, degree, and whether it is linear or non-linear:

(a) $x^2y'' - xy' + 5y = 0$

(b) $(y')^3 + 2y = x$

(c) $\frac{d^4y}{dx^4} + x\frac{d^3y}{dx^3} + x^2\frac{dy}{dx} = xe^x$

(d) $y'' - y' + y^3 = 3x$

(e) $(y'')^6 + y' = \cos x$

(f) $yy'' + x = 5$

(g) $y'' - xy' = xe^y$

Solution:

(a) Order: 2, Degree: 1, Linear

(b) Order: 1, Degree: 3, Non-linear

(c) Order: 4, Degree: 1, Linear

(d) Order: 2, Degree: 1, Non-linear (due to $y^3$)

(e) Order: 2, Degree: 6, Non-linear

(f) Order: 2, Degree: 1, Non-linear (due to $yy''$)

(g) Order: 2, Degree: 1, Non-linear (due to $e^y$)

Given a solution to a differential equation, we can find the corresponding differential equation by differentiating the given function and eliminating the arbitrary constants.

Example 6

Formulate the differential equation corresponding to each of the following solutions, where $A$ and $B$ are arbitrary constants:

(a) $y = A\sin x + B\cos x$

(b) $y = Ae^x + Be^{-2x}$

(c) $y = \frac{A}{x} + x$

Solution:

(a) Given $y = A\sin x + B\cos x$.

Differentiating with respect to $x$:

$\frac{dy}{dx} = A\cos x - B\sin x$

Differentiating again with respect to $x$:

$\frac{d^2y}{dx^2} = -A\sin x - B\cos x$

$\frac{d^2y}{dx^2} = -(A\sin x + B\cos x)$

$\frac{d^2y}{dx^2} = -y$

$\frac{d^2y}{dx^2} + y = 0$

(b) Given $y = Ae^x + Be^{-2x}$.

Differentiating with respect to $x$:

$\frac{dy}{dx} = Ae^x - 2Be^{-2x}$

Differentiating again with respect to $x$:

$\frac{d^2y}{dx^2} = Ae^x + 4Be^{-2x}$

Now we want to eliminate $A$ and $B$. From $y = Ae^x + Be^{-2x}$ and $\frac{dy}{dx} = Ae^x - 2Be^{-2x}$, we can derive:

$2y + \frac{dy}{dx} = 3Ae^x, \text{ so } A = \frac{2y + y'}{3e^x}$

$y - \frac{dy}{dx} = 3Be^{-2x}, \text{ so } B = \frac{y-y'}{3e^{-2x}}$

Substituting $A$ and $B$ into the second derivative equation:

$y'' = \frac{2y + y'}{3} + 4\frac{y-y'}{3}$

$3y'' = 2y + y' + 4y - 4y'$

$3y'' = 6y - 3y'$

$y'' + y' - 2y = 0$

(c) Given $y = \frac{A}{x} + x$.

Rewrite as $y = Ax^{-1} + x$.

Differentiating with respect to $x$:

$\frac{dy}{dx} = -Ax^{-2} + 1$

Differentiating again with respect to $x$:

$\frac{d^2y}{dx^2} = 2Ax^{-3}$

From the original equation, $A = x(y - x)$. Substituting this into the second derivative:

$\frac{d^2y}{dx^2} = 2(x(y - x))x^{-3} = \frac{2(y - x)}{x^2}$

$x^2\frac{d^2y}{dx^2} = 2y - 2x$

$x^2\frac{d^2y}{dx^2} - 2y + 2x = 0$

Example 7

Formulate the differential equation representing a circle of radius $r$ whose center is along the $y$-axis.

Solution:

The equation of a circle with center $(0, a)$ and radius $r$ is:

$x^2 + (y - a)^2 = r^2 \quad \text{(i)}$

Differentiating with respect to $x$:

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

$x + (y - a)\frac{dy}{dx} = 0$

$y - a = -\frac{x}{\frac{dy}{dx}}$

$y - a = -x\frac{dx}{dy} \quad \text{(ii)}$

Substitute (ii) into (i):

$x^2 + \left(-x\frac{dx}{dy}\right)^2 = r^2$

$x^2 + x^2\left(\frac{dx}{dy}\right)^2 = r^2$

$x^2\left(1 + \left(\frac{dx}{dy}\right)^2\right) = r^2$

Or, differentiating $x + (y - a)y' = 0$ again:

$1 + (y')^2 + (y-a)y'' = 0$

Substituting $y-a = -x/y'$:

$1 + (y')^2 - \frac{xy''}{y'} = 0$

$y'(1 + (y')^2) - xy'' = 0$

$y' + (y')^3 - xy'' = 0$