Question

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

Answer 1

Use the separation of variables method.

Explanation:

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`

Answer 2

`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)`