Explore the basic tenets of first and second order differential equations (linearity, degree, order, formulation, solutions, homogeneity, separability, and exactness)

The order of a differential equation is the highest derivative present in the equation. The degree is the exponent of the highest derivative after the equation has been cleared of radicals and fractions.

Examples

A differential equation is linear if the dependent variable and its derivatives occur only to the first degree, and no products of the dependent variable and its derivatives appear. The general form of an $n$th-order linear ordinary differential equation is:

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

A linear equation satisfies three conditions:

1. The dependent variable $y$ and all its derivatives occur to the first degree only
2. No product of $y$ and any of its derivatives are present
3. No transcendental functions (like $\sin y$, $\ln y$) of $y$ occur

Examples of Linear Differential Equations

• $\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + 3y = x^3$
• $\frac{dy}{dx} + e^xy = \ln x$

Examples of Non-linear Differential Equations

• $y\frac{dy}{dx} + 3x = 0$ (contains product $y \cdot y'$)
• $(y')^2 = 7xy$ (contains $(y')^2$)
• $\frac{dy}{dx} + x\sin y = 0$ (contains $\sin y$)

Example 1: For each differential equation, determine order, degree, and linearity.

(a) $xy'' - ye^x + 5 = 0$ — Order 2, Degree 1, Linear

(b) $(y')^3 + 2y = x$ — Order 1, Degree 3, Non-linear

(c) $\frac{d^4y}{dx^4} + x^2 \frac{d^3y}{dx^3} + x^3 \frac{dy}{dx} = xe^x$ — Order 4, Degree 1, Linear

To formulate a differential equation from a given solution:

1. Differentiate the given function as many times as the number of arbitrary constants present
2. Eliminate all arbitrary constants to obtain the differential equation

Example 2: Formulate the differential equation for $y = A\sin x + B\cos x$

Solution:

Given $y = A\sin x + B\cos x$

Differentiate with respect to $x$: $\frac{dy}{dx} = A\cos x - B\sin x$

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

Therefore, the required differential equation is: $\frac{d^2y}{dx^2} + y = 0$

Example 3: Formulate the differential equation for $y = Ae^x + Be^{-2x}$

Solution:

Given $y = Ae^x + Be^{-2x}$

Differentiating: $y' = Ae^x - 2Be^{-2x}$

$y - y' = (Ae^x + Be^{-2x}) - (Ae^x - 2Be^{-2x}) = 3Be^{-2x}$

$y' = y - 3Be^{-2x}$

Differentiating again: $y'' = y' + 2(3Be^{-2x}) = y' + 2(y - y')$

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

A solution is a function $y(x)$ whose derivatives satisfy the differential equation. Solutions can be verified by substituting derivatives into the original equation.

Verification Example

Verify that $f(x) = 2\sin x + 2\cos x$ is a solution of $f''(x) + f(x) = 0$

Solution:

$f(x) = 2\sin x + 2\cos x$

$f'(x) = 2\cos x - 2\sin x$

$f''(x) = -2\sin x - 2\cos x = -(2\sin x + 2\cos x) = -f(x)$

Therefore: $f''(x) + f(x) = -f(x) + f(x) = 0$

The equation is satisfied.

The general form is $y' = f(x, y)$ or $M(x, y)dx + N(x, y)dy = 0$

Separable Equations

A separable equation can be written as $\frac{dy}{dx} = g(x)h(y)$, which can be rearranged to: $\frac{dy}{h(y)} = g(x)dx$

Example 4: Solve $\frac{dy}{dx} = yx^2 + 3 + y + 3x^2$

Solution:

Rearrange: $\frac{dy}{dx} = (y+3)(x^2+1)$

Separate variables: $\frac{dy}{y+3} = (x^2+1)dx$

Integrate: $\int \frac{dy}{y+3} = \int (x^2+1)dx$

$\ln|y+3| = \frac{x^3}{3} + x + c$

$y = Ae^{\frac{1}{3}x^3+x} - 3$

Homogeneous Equations

A first order differential equation $M(x,y)dx + N(x,y)dy = 0$ is homogeneous if it can be written as $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$.

The substitution $u = \frac{y}{x}$ transforms it into a separable equation.

Example 5: Solve $\frac{dy}{dx} = \frac{y}{x} + 1$

Solution:

Let $u = \frac{y}{x}$, so $y = ux$

Then $\frac{dy}{dx} = u + x\frac{du}{dx}$

Substitute: $u + x\frac{du}{dx} = u + 1$

$x\frac{du}{dx} = 1$

Separate: $du = \frac{1}{x}dx$

Integrate: $u = \ln|x| + c$

Back-substitute: $\frac{y}{x} = \ln|x| + c$

$y = x\ln|x| + cx$

Exact Differential Equations

A differential equation $M(x,y)dx + N(x,y)dy = 0$ is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.

Example 6: Determine if $(x + 2y)dx + (y^2 + 2x)dy = 0$ is exact

Solution:

$M(x,y) = x + 2y$, so $\frac{\partial M}{\partial y} = 2$

$N(x,y) = y^2 + 2x$, so $\frac{\partial N}{\partial x} = 2$

Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact.

To solve: Find $F(x,y)$ such that $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$

$F(x,y) = \int (x + 2y)dx = \frac{x^2}{2} + 2xy + G(y)$

Differentiate with respect to $y$: $\frac{\partial F}{\partial y} = 2x + G'(y) = N = y^2 + 2x$

So $G'(y) = y^2$, and $G(y) = \frac{y^3}{3} + k$

Therefore: $\frac{x^2}{2} + 2xy + \frac{y^3}{3} = c$

The standard form is $\frac{dy}{dx} + P(x)y = Q(x)$

The integrating factor is $I(x) = e^{\int P(x)dx}$

Example 7: Solve $\frac{dy}{dx} + y = \sin x$

Solution:

Here $P(x) = 1$, so $I(x) = e^{\int 1dx} = e^x$

Multiply by $e^x$: $e^x\frac{dy}{dx} + e^x y = e^x \sin x$

$\frac{d}{dx}(ye^x) = e^x \sin x$

Integrate: $ye^x = \int e^x \sin x\, dx = \frac{1}{2}(e^x \sin x - e^x \cos x) + c$

$y = \frac{1}{2}(\sin x - \cos x) + ce^{-x}$

In Tanzania, differential equations are used in many practical situations. For example, when a farmer stores maize after harvest, the rate at which the moisture content decreases can be modeled by a differential equation similar to Newton's law of cooling. If the moisture content drops from 14% to 11% in the first week under ambient conditions of 25°C, a differential equation can predict when it will reach the safe storage level of 12.5%. This helps farmers and grain store operators in Arusha, Mbeya, and other agricultural regions determine optimal drying and storage times to prevent post-harvest losses.