This tutorial explains how to solve simple first-order differential equations of the form:
\[ \frac{dy}{dx} = f(x) \]These equations are solved by integrating both sides with respect to \( x \). The constant \( C \) represents the constant of integration.
Worked Examples
Step-by-Step Solved Examples
Example 1
Solve:
\[ \frac{dy}{dx} = 2x + 1 \]Solution:
\[ \int y' \, dx = \int (2x + 1)\,dx \] \[ y = x^2 + x + C \]You may verify by differentiation that this satisfies the original equation.
Example 2
Solve:
\[ 2\frac{dy}{dx} = \sin(2x) \]Solution:
\[ y' = \frac{1}{2}\sin(2x) \] \[ y = \int \frac{1}{2}\sin(2x)\,dx \]Let \( u = 2x \), so \( du = 2dx \implies dx = \frac{1}{2}du \):
\[ y = \int \frac{1}{4}\sin(u)\,du \] \[ y = -\frac{1}{4}\cos(u) + C = -\frac{1}{4}\cos(2x) + C \]Example 3
Solve:
\[ y'e^{-x} + e^{2x} = 0 \]Solution:
Multiply both sides by \( e^x \), simplify and rewrite as:
\[ y' = -e^{3x} \]Integrate:
\[ y = \int -e^{3x}\,dx \]Let \( u = 3x \), \( du = 3dx \):
\[ y = \int -\frac{1}{3}e^{u}\,du \] \[ y = -\frac{1}{3}e^{3x} + C \]Practice Exercises and Answers
Practice Problems & Solutions
Solve the following:
- \( 2\dfrac{dy}{dx} = 6x \)
- \( y'\cos(x) = \sin(2x) \)
- \( y'e^x = e^{3x} \)
Answers:
- \( y = \dfrac{3}{2}x^2 + C \)
- \( y = -2\cos(x) + C \)
- \( y = \dfrac{1}{2}e^{2x} + C \)