Question

What is the general solution of the differential equation? : `dy/dx = x+2y`

Answer 1

The General Solution is:

`y = -1/2x -1/4 + Ce^(2x)`

Explanation:

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

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

We have:

`dy/dx = x+2y`

Which we can write as:

`dy/dx -2y = x` ..... [A]

This is a First Order Ordinary Differential Equation in Standard Form. So we compute and integrating factor, `I`, using;

`I = e^(int P(x) dx)`
`\ \ = exp(int \ -2 \ dx)`
`\ \ = exp( -2x)`
`\ \ = e^(-2x)`

And if we multiply the DE [A] by this Integrating Factor, `I`, we will have a perfect product differential;

`dy/dxe^(-2x) -2ye^(-2x) = xe^(-2x)`

`:. d/dx( ye^(-2x) ) = xe^(-2x)`

This has transformed our initial ODE into a Separable ODE, so we can now "separate the variables" to get::

`ye^(-2x) = int \ xe^(-2x) \ dx`

We can proceed via an application of integration by Parts

Let `{ (u,=x, => (du)/dx,=1), ((dv)/dx,=e^(-2x), => v,=-1/2e^(-2x) ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

We have:

`int \ (x)(e^(-2x)) \ dx = (x)(-1/2e^(-2x)) - int \ (-1/2e^(-2x))(1) \ dx`
`:. int \ xe^(-2x) \ dx = -1/2xe^(-2x) +1/2 \ int \ e^(-2x) \ dx`
`" " = -1/2xe^(-2x) -1/4 e^(-2x) + C`

Using this result, we can write the DE Solution as :

`ye^(-2x) = -1/2xe^(-2x) -1/4 e^(-2x) + C`

`:. y = -1/2x -1/4 + Ce^(2x)`