Solving First-Order Differential Equations

Introduction

Key Concepts

1. Understanding Differential Equations

A differential equation is an equation that relates a function with its derivatives. First-order differential equations involve the first derivative of the function and can be expressed in the general form:

$$\frac{dy}{dx} = f(x, y)$$

These equations are pivotal in modeling dynamic systems where the rate of change of a quantity depends on its current state and another variable, typically time or space.

2. Classification of First-Order Differential Equations

First-order differential equations can be classified based on their form and the methods required for their solutions. The primary classifications include:

• Separable Equations: These can be written as $g(y) dy = f(x) dx$, allowing the variables to be separated on either side of the equation.
• Linear Equations: These take the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are continuous functions.
• Exact Equations: These satisfy the condition $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, where the equation is written as $M(x, y) dx + N(x, y) dy = 0$.
• Integrating Factor Method: Used for linear equations, an integrating factor $\mu(x)$ is employed to simplify the equation into an exact differential.

3. Solving Separable Equations

Separable equations are the most straightforward to solve. The steps involved include:

1. Rewrite the equation to separate the variables:

$$\frac{dy}{dx} = g(x)h(y) \Rightarrow \frac{1}{h(y)} dy = g(x) dx$$

2. Integrate both sides:

$$\int \frac{1}{h(y)} dy = \int g(x) dx$$

3. Solve for $y$ to obtain the general solution.

Example: Solve $\frac{dy}{dx} = xy$.

Solution:

Separate variables:

$$\frac{1}{y} dy = x dx$$

Integrate both sides:

$$\ln|y| = \frac{x^2}{2} + C$$

Exponentiate to solve for $y$:

$$y = Ce^{\frac{x^2}{2}}$$

4. Solving Linear Differential Equations

Linear differential equations can be solved using the integrating factor method. The standard form is:

$$\frac{dy}{dx} + P(x)y = Q(x)$$

The integrating factor $\mu(x)$ is calculated as:

$$\mu(x) = e^{\int P(x) dx}$$

Multiply both sides of the differential equation by $\mu(x)$:

$$\mu(x)\frac{dy}{dx} + \mu(x)P(x)y = \mu(x)Q(x)$$

This simplifies to:

$$\frac{d}{dx} [\mu(x)y] = \mu(x)Q(x)$$

Integrate both sides to find $y$:

$$y = \frac{1}{\mu(x)} \left( \int \mu(x)Q(x) dx + C \right)$$

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

Solution:

Identify $P(x) = 2$ and $Q(x) = e^{3x}$.

Calculate the integrating factor:

$$\mu(x) = e^{\int 2 dx} = e^{2x}$$

Multiply the differential equation by $\mu(x)$:

$$e^{2x}\frac{dy}{dx} + 2e^{2x}y = e^{5x}$$

The left side becomes the derivative of $e^{2x}y$:

$$\frac{d}{dx} (e^{2x}y) = e^{5x}$$

Integrate both sides:

$$e^{2x}y = \frac{e^{5x}}{5} + C$$

Solve for $y$:

$$y = \frac{e^{3x}}{5} + Ce^{-2x}$$

5. Exact Differential Equations

An exact differential equation has the form $M(x, y) dx + N(x, y) dy = 0$ and satisfies the condition:

$$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$

If the equation is exact, there exists a potential function $\Psi(x, y)$ such that:

$$\Psi(x, y) = C$$

To find $\Psi(x, y)$:

1. Integrate $M(x, y)$ with respect to $x$:

$$\Psi(x, y) = \int M(x, y) dx + h(y)$$

2. Differentiate $\Psi(x, y)$ with respect to $y$ and set it equal to $N(x, y)$:

$$\frac{\partial \Psi}{\partial y} = N(x, y)$$

3. Solve for $h(y)$ and substitute back into $\Psi(x, y)$.

Example: Solve $(2xy + y^2) dx + (x^2 + 2xy) dy = 0$.

Solution:

Identify $M = 2xy + y^2$ and $N = x^2 + 2xy$.

Check exactness:

$$\frac{\partial M}{\partial y} = 2x + 2y$$

$$\frac{\partial N}{\partial x} = 2x + 2y$$

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

Find $\Psi(x, y)$:

Integrate $M$ with respect to $x$:

$$\Psi(x, y) = \int (2xy + y^2) dx = x^2 y + y^2 x + h(y)$$

Differentiate with respect to $y$ and set equal to $N$:

$$\frac{\partial \Psi}{\partial y} = x^2 + 2xy + h'(y) = x^2 + 2xy$$

Thus, $h'(y) = 0 \Rightarrow h(y) = C$.

Therefore, the general solution is:

$$x^2 y + y^2 x = C$$

6. Applications of First-Order Differential Equations

First-order differential equations are instrumental in modeling various real-life situations, including:

• Population Dynamics: Modeling population growth or decay using the logistic equation.
• Chemical Reactions: Describing the rate of reaction between reactants.
• Thermodynamics: Modeling cooling and heating processes.
• Finance: Calculating interest rates and investment growth.

Example: Newton's Law of Cooling states that the rate of cooling of an object is proportional to the difference in temperatures between the object and the surrounding environment.

The differential equation is:

$$\frac{dT}{dt} = -k(T - T_s)$$

Where $T(t)$ is the temperature of the object at time $t$, $T_s$ is the surrounding temperature, and $k$ is a positive constant.

Solving this equation using the integrating factor method yields:

$$T(t) = T_s + (T_0 - T_s)e^{-kt}$$

Where $T_0$ is the initial temperature of the object.

7. Integrating Factors and Their Role

An integrating factor is a function that simplifies a linear differential equation, making it exact and easier to solve. It is essential in transforming the original equation into a form where the left side becomes the derivative of a product of functions.

The integrating factor $\mu(x)$ is determined by:

$$\mu(x) = e^{\int P(x) dx}$$

Multiplying the entire differential equation by $\mu(x)$ ensures that the equation can be integrated directly to find the solution.

8. Initial Conditions and Particular Solutions

Initial conditions provide specific values of the function and its derivatives at a particular point, allowing for the determination of the constant of integration in the general solution. Applying an initial condition transforms the general solution into a particular solution tailored to the given scenario.

Example: Given the differential equation $\frac{dy}{dx} = 3y$, with the initial condition $y(0) = 2$, find the particular solution.

Solution:

Separate variables:

$$\frac{1}{y} dy = 3 dx$$

Integrate both sides:

$$\ln|y| = 3x + C$$

Solve for $y$:

$$y = Ce^{3x}$$

Apply the initial condition $y(0) = 2$:

$$2 = Ce^{0} \Rightarrow C = 2$$

Thus, the particular solution is:

$$y = 2e^{3x}$$

9. Homogeneous and Non-Homogeneous Equations

First-order differential equations can be homogeneous or non-homogeneous based on whether the equation can be expressed in terms of a single homogeneous function.

• Homogeneous Equations: These can be written as $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$.
• Non-Homogeneous Equations: These cannot be expressed solely in terms of $\frac{y}{x}$ and may include additional functions of $x$ or $y$.

Solution methods vary accordingly, with homogeneous equations often being solvable through substitution, while non-homogeneous equations may require integrating factors or other techniques.

10. Bernoulli's Equation

Bernoulli's equation is a special type of nonlinear first-order differential equation of the form:

$$\frac{dy}{dx} + P(x)y = Q(x)y^n$$

Where $n$ is a real number. To solve Bernoulli's equation, a substitution is made to transform it into a linear differential equation. Letting $v = y^{1-n}$, the equation becomes linear in terms of $v$ and can be solved using the integrating factor method.

Example: Solve $\frac{dy}{dx} + y = y^2$.

Solution:

Identify $P(x) = 1$, $Q(x) = 1$, and $n = 2$.

Substitute $v = y^{1-2} = y^{-1}$.

Differentiate $v$ with respect to $x$:

$$\frac{dv}{dx} = -y^{-2}\frac{dy}{dx}$$

Rewrite the original equation in terms of $v$:

$$\frac{dy}{dx} + y = y^2 \Rightarrow \frac{dy}{dx} = y^2 - y$$

Substitute into the derivative of $v$:

$$\frac{dv}{dx} = -y^{-2}(y^2 - y) = -1 + y^{-1} = -1 + v$$

The equation becomes:

$$\frac{dv}{dx} - v = -1$$

This is a linear differential equation. The integrating factor is:

$$\mu(x) = e^{-\int 1 dx} = e^{-x}$$

Multiply both sides by $\mu(x)$:

$$e^{-x}\frac{dv}{dx} - e^{-x}v = -e^{-x}$$

The left side simplifies to:

$$\frac{d}{dx}(e^{-x}v) = -e^{-x}$$

Integrate both sides:

$$e^{-x}v = e^{-x} + C$$

Solve for $v$:

$$v = 1 + Ce^{x}$$

Recall that $v = y^{-1}$:

$$y = \frac{1}{1 + Ce^{x}}$$

Comparison Table

Summary and Key Takeaways