The document provides an overview of linear first-order differential equations, detailing their structure and general solution. It includes several examples with step-by-step solutions for specific differential equations, demonstrating the use of integrating factors and initial conditions. Key results show how to derive particular solutions based on given conditions.

1. Differential Equation
Name: Ahmad Anwar
Hassan Shah

2. Linear First Order
• A linear first-order differential equation is a differential equation of
the form:
• y is the unknown function
• x is the independent variable
• P(x) and Q(x) are functions of x
• dy/dx is the derivative of y with respect to x

3. The general solution of the differential equation is:
• y = (1/e^∫P(x)dx) ∫(Q(x)e^∫P(x)dx)dx + Ce^(-∫P(x)dx)
• where C is the constant of integration.
• Here are some examples:

4. Solve the differential equation: dy/dx + 2y = 3
• Solution:
• Integrating factor = e^∫2dx = e^(2x)
• Multiply both sides by the integrating factor:
• e^(2x)dy/dx + 2e^(2x)y = 3e^(2x)
• Now, integrate both sides:
• ∫(e^(2x)dy) + ∫(2e^(2x)y)dx = ∫(3e^(2x))dx
• e^(2x)y = (3/2)e^(2x) + C
• y = (3/2) + Ce^(-2x)

5. Solve the differential equation: dy/dx - 3y = 2
• Solution:
• Integrating factor = e^∫(-3)dx = e^(-3x)
• Multiply both sides by the integrating factor:
• e^(-3x)dy/dx - 3e^(-3x)y = 2e^(-3x)
• Now, integrate both sides:
• ∫(e^(-3x)dy) - ∫(3e^(-3x)y)dx = ∫(2e^(-3x))dx
• e^(-3x)y = (-2/3)e^(-3x) + C
• y = (-2/3) + Ce^(3x)

6. Numerical Problems
• Solve the differential equation: dy/dx + 4y = 5, with initial condition y(0) = 1
• Solution:
• Integrating factor = e^∫4dx = e^(4x)
• Multiply both sides by the integrating factor:
• e^(4x)dy/dx + 4e^(4x)y = 5e^(4x)
• Now, integrate both sides:
• ∫(e^(4x)dy) + ∫(4e^(4x)y)dx = ∫(5e^(4x))dx
• e^(4x)y = (5/4)e^(4x) + C
• Using the initial condition y(0) = 1, we get:
• 1 = (5/4) + C
• C = -1/4
• y = (5/4)e^(-4x) + (1/4)e^(-4x)
• y = (1/4)(5e^(-4x) + 1)

7. Solve the differential equation: dy/dx - 2y = 3x, with
initial condition y(1) = 2
• Solution:
• Integrating factor = e^∫(-2)dx = e^(-2x)
• Multiply both sides by the integrating factor:
• e^(-2x)dy/dx - 2e^(-2x)y = 3xe^(-2x)
• Now, integrate both sides:
• ∫(e^(-2x)dy) - ∫(2e^(-2x)y)dx = ∫(3xe^(-2x))dx
• e^(-2x)y = (-3/2)xe^(-2x) + (3/2)e^(-2x) + C
• Using the initial condition y(1) = 2, we get:
• 2 = (-3/2)e^(-2) + (3/2)e^(-2) + C
• C = 2e^2
• y = (-3/2)xe^(-2x) + (3/2)e^(-2x) + 2e^(2-2x)
• y = (-3/2)xe^(-2x) + (1/2)e^(-2x) + 2e^(-2x)
• Note: e^(2-2x) = e^(2)e^(-2x) = e^2e^(-2x)