Question

What is the general solution of the differential equation? : `dy/dx + 3y = 3x^2e^(-3x)`

Answer 1

`y=(x^3+C)*e^(-3x)`

Explanation:

`dy/dx+3y=3x^2*e^(-3x)`

`dy/dx*e^(3x)+3y*e^(3x)=3x^2*e^(-3x)*e^(3x)`

`d/dx (y*e^(3x))=3x^2`

`y*e^(3x)=x^3+C`

`y=(x^3+C)*e^(-3x)`

Note: This differential equation is first order and linear one.

1) I multiplied both sides with `e^(3x)` for converting left side to the exact differential equation.

2) I integrated both sides.

3) I multiplied with `e^(-3x)` for solving `y`.

Answer 2

`y = x^3e^(-3x) + Ce^(-3x)`

Explanation:

We have:

`dy/dx + 3y = 3x^2e^(-3x)` ..... [1]

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)`

Then the integrating factor is given by;

`I = e^(int P(x) dx)`
`\ \ = exp(int \ 3 \ dx)`
`\ \ = exp( 3x )`
`\ \ = e^(3x)`

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

`e^(3x)dy/dx + 3e^(3x)y = 3x^2e^(-3x)e^(3x)`

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

Which we can now directly integrate to get:

`e^(3x)y = int \ x^3 \ dx`

`:. e^(3x)y = x^3+ C`

`:. y = (x^3+ C)/e^(3x)`

`:. y = x^3e^(-3x) + Ce^(-3x)`