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

First we notice that

$d \frac{\sin {x}^{2}}{\mathrm{dx}} = 2 x \cos {x}^{2}$

or

$x \cos {x}^{2} = \frac{1}{2} \cdot \left(\mathrm{ds} \in {x}^{2} / \mathrm{dx}\right)$

Hence the problem becomes

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \cdot \frac{\mathrm{ds} \in {x}^{2}}{\mathrm{dx}}$

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

$\int \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \mathrm{dx} = \frac{1}{2} \cdot \int \left(\mathrm{ds} \in {x}^{2} / \mathrm{dx}\right) \mathrm{dx}$

$y = \frac{1}{2} \cdot \sin {x}^{2} + c$

The general solution is

$y \left(x\right) = \frac{1}{2} \cdot \sin {x}^{2} + c$

Mar 5, 2017

$y = \frac{1}{2} \sin \left({x}^{2}\right) + C$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \cos {x}^{2}$

$\mathrm{dy} = x \cos {x}^{2} \mathrm{dx}$

$\int \mathrm{dy} = \int x \cos {x}^{2} \mathrm{dx}$

THIS SOLUTION IS ONLY CORRECT IF THE PROBLEM IS WRITTEN CORRECTLY. The solution would be different if the problem is $\frac{\mathrm{dy}}{\mathrm{dx}} = x {\cos}^{2} x$.

Let $u = {x}^{2}$. Then $\mathrm{du} = 2 x \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{2 x}$.

$\int \mathrm{dy} = \int x \cos u \frac{\mathrm{du}}{2 x}$

$\int \mathrm{dy} = \int \frac{1}{2} \cos u \mathrm{du}$

$y = \frac{1}{2} \sin u + C$

$y = \frac{1}{2} \sin \left({x}^{2}\right) + C$

Hopefully this helps!