Simple But Practical: First-Order Ordinary Differential Equations

Examples of first-order differential equations:

1. Linear ODE: \[ \frac{dy}{dx} + y = x \]
   • A first-order linear differential equation.
2. Separable ODE: \[ \frac{dy}{dx} = xy \]
   • A separable differential equation, where variables can be separated on each side of the equation.
3. Exact ODE: \[ (2xy + e^y)\,dx + (x^2 + xe^y)\,dy = 0 \]
   • An exact differential equation, characterized by its form involving differentials \( dx \) and \( dy \).
   • Exact differential equations are a type of equation where combining two related functions correctly gives a solution. They’re solved by finding a special function that ties these two together.
4. Homogeneous ODE: \[ \frac{dy}{dx} = \frac{y – x}{y + x} \]
   • A homogeneous differential equation, where the function is homogeneous with respect to \( x \) and \( y \).
   

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5. Bernoulli’s Equation: \[ \frac{dy}{dx} + p(x)y = q(x)y^n \]
   • Bernoulli’s equation, a special case of a non-linear differential equation, where \( p(x) \) and \( q(x) \) are continuous functions of \( x \), and \( n \) is a real number.

Solving first order ode:

To solve a first-order ODE (ordinary differential equation), identify the type, such as separable, linear, exact, or integrable through substitution. Each type has its own characteristic structure and corresponding solution method. Identifying the type helps in choosing the right approach to solve the equation. For additional educational resources,.

Let’s look at this example. It is solved using the principles of separable differential equation:

Problem:

Consider the separable differential equation \(\frac{dy}{dx} = \frac{x}{y}, \text{ where } y \neq 0.\)

Solution:

Step 1: Separate Variables. Rearrange the equation to separate \(x \text{ and } y\):

\(y\, dy = x\, dx.\)

Step 2: Integrate Both Sides. Integrate both sides of the equation:

\(\int y\, dy = \int x\, dx.\)

Step 3: Perform the Integration. Carrying out the integration gives:

\(\frac{y^2}{2} = \frac{x^2}{2} + C\),

where \(C\) is the constant of integration.

Step 4: Solve for \(y\).

Rearrange the equation to solve for \(y\):

\(y^2 = x^2 + 2C\),

\(y = \pm \sqrt{x^2 + 2C}\)

Thus, the general solution to the differential equation \(\frac{dy}{dx} = \frac{x}{y}\) is:

\(y = \pm \sqrt{x^2 + 2C}\),

\(C\) is a constant determined by initial conditions.

Here is an example regarding exact differential equations:

Problem:

Solve the exact differential equation given by

\(\left( 2y – 3x^2 \right)dx + \left( 2x – 4y \right)dy = 0.\)

Solution:

First, verify if the equation is exact. For an equation of the form

\(M(x, y)dx + N(x, y)dy = 0\), it is exact if

\(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}.\)

Here, \(M(x, y) = 2y – 3x^2 \text{ and } N(x, y) = 2x – 4y.\)

\(\frac{\partial M}{\partial y} = 2, \quad \frac{\partial N}{\partial x} = 2.\)

Since these are equal, the differential equation is exact.

To find the potential function \(\Psi(x, y)\), integrate \(M\) with respect to \(x\)

and \(N\) with respect to \(y\):

\(\int (2y – 3x^2) dx = 2yx – x^3 + h(y)\),

\(\int (2x – 4y) dy = 2xy – 2y^2 + g(x).\)

Compare the two integrals to determine \(h(y)\) and \(g(x)\):

\(\Psi(x, y) = 2yx – x^3 – 2y^2 + C\),

where \(C\) is a constant.

The general solution to the differential equation is given by the level curves of \(\Psi(x, y)\):

\(2yx – x^3 – 2y^2 = C. \)