Question

What is the general solution of the differential equation `y'-2xy=x^3`?

Answer 1

`y = 5/2e^( x^2 )-1/2x^2-1/2`

Explanation:

We have:

`y'-2xy=x^3` ..... [1]

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

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

We can readily generate an integrating factor when we have an equation of this form, given by;

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

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

`e^( -x^2 )y'-2xe^( -x^2 )y=x^3e^( -x^2 )`
`:. d/dx(e^( -x^2 )y) = x^3e^( -x^2 )`

We can now integrate to get:

`e^( -x^2 )y = int \ x^3e^( -x^2 ) \ dx + C`

The RHS integral can be evaluated by an application of Integration By Parts (omitted) which gives us:

`int \ x^3e^( -x^2 ) \ dx = -1/2(x^2+1)e^(-x^2)`

So we have:

`e^( -x^2 )y = -1/2(x^2+1)e^(-x^2) + C`

Using the initial condition `y(0)=2` we can evaluate the constant `C`

`2e^0 = -1/2(0+1)e^0 + C => C = 5/2`

Leading to the Particular Solution:

`e^( -x^2 )y = -1/2(x^2+1)e^(-x^2) +5/2`

`:. y = -1/2(x^2+1)e^(-x^2) +5/(2e^( -x^2 ))`
`\ \ \ \ = -1/2(x^2+1) +5/2e^( x^2 )`
`\ \ \ \ = 5/2e^( x^2 )-1/2x^2-1/2`