To solve this differential equation, rewrite it as

`dy = xcos(x^2)dx`

Integrate both sides of the equation:

`y = int xcos(x^2)dx`

To take the integral on the right side of the equation, use the substitution method. Let `z(x) = x^2` .

Then, `dz = 2xdx` and integral becomes

`int xcos(x^2)dx = int(xdx) cos(x^2) = int 1/2 dz cosz = 1/2 int coszdz`

This is a simple trigonometric integral: `int cosz dz = sinz` . Substituting the original variable, x, back into equation results in

`y = 1/2 sinx^2 + C` , where C is an arbitrary constant.

**So, the general solution of the given differential equation
is**

`y = 1/2sinx^2 + C` .

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