Question

How can i solve this differencial equation? : `y'+x^2 y=x^2`

Answer 1

`y = 1 + Ce^(-1/3x^3)`

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:

`y'+x^2y=x^2` ..... [1]

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 \ x^2 \ dx)`
`\ \ = exp( 1/3x^3 )`
`\ \ = e^(1/3x^3)`

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

`y'e^(1/3x^3)+x^2e^(1/3x^3)y = x^2e^(1/3x^3)`

`:. d/dx( ye^(1/3x^3) ) = x^2e^(1/3x^3)`

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

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

This is trivial to integrate, and we get:

`ye^(1/3x^3) = e^(1/3x^3) + C`

Leading to the explicit General Solution:

`y = e^(-1/3x^3){e^(1/3x^3) + C}`
`\ \ = 1 + Ce^(-1/3x^3)`

Answer 2

A simpler method to the one given in the other answer

Explanation:

`y'+x^2y=x^2`

This is a separable ODE

`dy/dx=x^2-x^2y=x^2(1-y)`

`rArr int1/(1-y)dy=intx^2dx`

`rArr-ln|1-y|=1/3x^3+"c"`

`1-y=Be^(1/3x^3)` where `B=e^(-"c")`

`y=1+Ae^(1/3x^3)` where `A=-B`