# What is the general solution of the differential equation # dy/dx - 2y + a = 0 #?

##### 2 Answers

Use the separation of variables method.

#### Explanation:

Given:

Add

Multiply both sides by

Integrate both sides:

Multiply both sides by 2:

Use the exponential function on both sides:

The inverses on the left disappear:

Adding an arbitrary constant in the exponent is the same a multiplying by an arbitrary constant:

Add

# y = 1/2 a +Ce^(2x) #

#### Explanation:

First write the DE in standard form:

# dy/dx - 2y + a = 0 #

# :. dy/dx - 2y = - a # ... [1]

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

This is a standard form of a Differential Equation that can be solved by using an Integrating Factor:

# I = e^(int P(x) dx)#

# \ \ = e^(int \ -2 \ dx)#

# \ \ = e^(-2x) #

And if we multiply the DE [1] by this Integrating Factor we will have a perfect product differential;

# dy/dx e^(-2x)- 2ye^(-2x) = - ae^(-2x) #

# d/dx(ye^(-2x)) = - ae^(-2x) #

This has converted our DE into a First Order separable DE which we can now just separate the variables to get;

# ye^(-2x) = int \ - ae^(-2x) \ dx #

Which we can easily integrate to get:

# ye^(-2x) = 1/2 ae^(-2x) +C #

# :. y = 1/2 a +Ce^(2x) #