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

Question

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

William Abdalla · Accepted Answer

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

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) #

Emilia Aguilar · Answer

Use the separation of variables method.

Given: #dy/dx - 2y + a = 0# Add #2y - a# to both sides: #dy/dx = 2y - a # Multiply both sides by #dx/(2y-a)#: #dy/(2y - a)=dx # #1/2dy/(y-a/2)=dx# Integrate both sides: #1/2intdy/(y-a/2)=intdx# #1/2ln(y-a/2)=x+C# Multiply both sides by 2: #ln(y-a/2)=2x+C# Use the exponential function on both sides: #e^(ln(y-a/2))=e^(2x+C)# The inverses on the left disappear: #y-a/2=e^(2x+C)# Adding an arbitrary constant in the exponent is the same a multiplying by an arbitrary constant: #y-a/2=Ce^(2x)# Add #a/2# to both sides: #y=Ce^(2x)+a/2#

Ezra Adams · Answer

undefined The general solution of the differential equation $ \frac{dy}{dx} - 2y + a = 0 $ is $ y = \frac{a}{2} + Ce^{2x} $, where $ C $ is an arbitrary constant.