Basic Differential Equations

These are the most straightforward types of differential equations: $$ f'(x) = g(x) $$ They can be solved directly by integration: $$ f(x) = \int g(x)\, dx = F(x) + c $$

A Simple Example

Consider the elementary differential equation:

$$ f'(x) = 2x $$

To solve it, we integrate both sides with respect to \( x \):

$$ \int f'(x)\, dx = \int 2x\, dx $$

$$ f(x) = x^2 + c $$

This gives the general solution to the equation.

Example 2

Let’s look at a slightly more involved case:

$$ f'(x) = 3e^{2x} $$

To find the unknown function \( f(x) \), we again integrate both sides:

$$ \int f'(x)\, dx = \int 3e^{2x}\, dx $$

Since the integral of the derivative returns the original function:

$$ f(x) = \int 3e^{2x}\, dx $$

We can factor out the constant:

$$ f(x) = 3 \cdot \int e^{2x}\, dx $$

To evaluate the integral, we use the substitution rule or recognize that \( e^{2x} \) integrates to \( \frac{1}{2}e^{2x} \):

$$ f(x) = 3 \cdot \frac{1}{2}e^{2x} + c = \frac{3}{2}e^{2x} + c $$

So the general solution is: $$ f(x) = \frac{3e^{2x}}{2} + c $$

Example 3

This is a second-order differential equation, but it’s still considered basic because it can be solved with two successive integrations:

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

We start by integrating once:

$$ \int f''(x)\, dx = \int (2 - \cos x)\, dx $$

$$ f'(x) = 2x - \sin x + c_1 $$

This yields the first derivative of the solution. We now integrate again:

$$ \int f'(x)\, dx = \int (2x - \sin x + c_1)\, dx $$

$$ f(x) = x^2 + \cos x + c_1x + c_2 $$

So the general solution is: \( f(x) = x^2 + \cos x + c_1x + c_2 \).

And so on.