Question

Solve the differential equation:

`(1 + x^2) dy/dx = 4x^2 - 2xy`

Answer 1

Given differential equation is,

`(1 + x^2) dy/dx = 4x^2 - 2xy`

⇒ `dy/dx = (4x^2)/(1 + x^2) - (2xy)/(1 + x^2)`

or `dy/dx + (2xy)/(1 + x^2) = (4x^2)/(1 + x^2)`

This is of the form `dy/dx + "P"y = "Q"`

where, P = `(2x)/(1 + x^2)` and Q = `(4x^2)/(1 + x^2)`

Thus, the solution of given differential equation is,

`y xx "I.F." = int ("Q" xx "I.F.") dx + c`

where, I.F. = `e^(int P dx)`

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

= `e^(log(1 + x^2))`      ...`[∵ int (f'(x))/(f(x)) dx = log [f(x)] + c]`

= 1 + x²   ...`[∵ e^(log(a)) = a]`

∴ `y xx (1 + x^2) = int[(4x^2)/(1 + x^2) xx (1 + x^2)] dx + c`

⇒ `(1 + x^2)y = int 4x^2 dx + c`

⇒ `(1 + x^2)y = (4x^3)/3 + c`