Question

What is the solution of the Differential Equation `x(dy/dx)+3y+2x^2=x^3+4x`?

Answer 1

`y = (5x^6 - 12x^5+30x^4 + C)/(30x^3)`

Explanation:

We have:

`x(dy/dx)+3y+2x^2=x^3+4x`

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)`

So rewrite the equations in standard form as:

`(dy/dx)+3/xy = x^2-2x+4 ..... [1]`

Then the integrating factor is given by;

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

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

`(dy/dx)x^3+3x^2y = x^5-2x^4+4x^3`

`:. d/dx (x^3y) = x^5-2x^4+4x^3`

Which we can directly integrate to get:

`x^3y = int \ x^5-2x^4+4x^3 \ dx`

`:. x^3y = x^6/6-(2x^5)/5+x^4 + c`

`:. x^3y = (5x^6 - 12x^5+30x^4 + 30c)/30`

`:. y = (5x^6 - 12x^5+30x^4 + C)/(30x^3)`

Confirming the given answer.