How to you find the general solution of $\frac{d y}{d x} = x {cos x}^{2}$ $dy/dx=xcosx^2$ ?

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How to you find the general solution of $\frac{d y}{d x} = x {cos x}^{2}$ $dy/dx=xcosx^2$ ?

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Answer 1 · 2017-03-05T09:40:11

First we notice that

$d \frac{{sin x}^{2}}{d x} = 2 x {cos x}^{2}$ $d(sinx^2)/dx=2xcosx^2$

or

$x {cos x}^{2} = \frac{1}{2} \cdot (d \frac{{sin x}^{2}}{d x})$ $xcosx^2=1/2*(dsinx^2/dx)$

Hence the problem becomes

$\frac{d y}{d x} = \frac{1}{2} \cdot \frac{d {sin x}^{2}}{d x}$ $dy/dx=1/2*(dsinx^2)/dx$

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

$\int \frac{d y}{d x} \cdot d x = \frac{1}{2} \cdot \int (d \frac{{sin x}^{2}}{d x}) d x$ $int dy/dx*dx =1/2*int (dsinx^2/dx)dx$

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

The general solution is

$y (x) = \frac{1}{2} \cdot {sin x}^{2} + c$ $y(x)=1/2*sinx^2+c$

Answer 2 · 2017-03-05T15:54:16

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

Explanation:

$\frac{d y}{d x} = x {cos x}^{2}$ $dy/dx = xcosx^2$

$d y = x {cos x}^{2} d x$ $dy = xcosx^2dx$

$\int d y = \int x {cos x}^{2} d x$ $int dy = int xcosx^2 dx$

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

Let $u = x^{2}$ $u = x^2$ . Then $d u = 2 x d x$ $du = 2xdx$ and $d x = \frac{d u}{2 x}$ $dx = (du)/(2x)$ .

$\int d y = \int x cos u \frac{d u}{2 x}$ $intdy = intxcosu (du)/(2x)$

$\int d y = \int \frac{1}{2} cos u d u$ $intdy = int 1/2cosudu$

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

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

Hopefully this helps!

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How to you find the general solution of $\frac{d y}{d x} = x {cos x}^{2}$ $dy/dx=xcosx^2$ ?

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