Question

What is the general solution of the differential equation `y'-1/xy=2x^2\lnx`?

(because Symbolab is addicted to weird `C` constants...)

Answer 1

`y = x^3lnx - x^3/2 + Cx`

Explanation:

We have:

`y'-1/x y=2x^2 lnx`

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)`

As the equation is already in this form, then the integrating factor is given by;

`I = e^(int P(x) dx)`
`\ \ = exp(int \ -1/x \ dx)`
`\ \ = exp(-lnx)`
`\ \ = exp(ln(1/x))`
`\ \ = 1/x`

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

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

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

`:. d/dx (y/x) = 2x lnx`

This is now separable, so by "separating the variables" we get:

`y/x = 2 \ int \ x lnx \ dx`

To integrate the RHS integral we can apply integration by Parts:

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

Then plugging into the IBP formula:

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

We have:

`int \ (lnx)(x) \ dx = (lnx)(x^2/2) - int \ (x^2/2)(1/x) \ dx`

`:. int \ xlnx \ dx = (x^2lnx)/2 - int \ x/2 \ dx`
`" " = (x^2lnx)/2 - x^2/4`

With this result, we have:

`y/x = 2 {(x^2lnx)/2 - x^2/4} + C`

Leading to the GS:

`y = 2x {(x^2lnx)/2 - x^2/4} + Cx`

`\ \ = x^3lnx - x^3/2 + Cx`