Question

How to you find the general solution of `dy/dx=xcosx^2`?

Answer 1

First we notice that

`d(sinx^2)/dx=2xcosx^2`

or

`xcosx^2=1/2*(dsinx^2/dx)`

Hence the problem becomes

`dy/dx=1/2*(dsinx^2)/dx`

Integrate both sides with respect to `x` and we have

`int dy/dx*dx =1/2*int (dsinx^2/dx)dx`

`y=1/2*sinx^2+c`

The general solution is

`y(x)=1/2*sinx^2+c`

Answer 2

`y = 1/2sin(x^2)+ C`

Explanation:

`dy/dx = xcosx^2`

`dy = xcosx^2dx`

`int dy = int xcosx^2 dx`

THIS SOLUTION IS ONLY CORRECT IF THE PROBLEM IS WRITTEN CORRECTLY. The solution would be different if the problem is `dy/dx = xcos^2x`.

Let `u = x^2`. Then `du = 2xdx` and `dx = (du)/(2x)`.

`intdy = intxcosu (du)/(2x)`

`intdy = int 1/2cosudu`

`y = 1/2sinu + C`

`y = 1/2sin(x^2) + C`

Hopefully this helps!