How to you find the general solution of dydx=xcos⁡x2?

dydx=xcos⁡x2dy=xcos⁡x2dx∫dy=∫xcos⁡x2dxLet u=x2. Then du=2xdxanddx=du2x∫dy=∫xcos⁡udu2x∫dy=∫12cos⁡uduy=12sin⁡u+Cy=12sin⁡(x2)+C

Step by step solution to "How to you find the general solution of..."

Alfredo Holmes

2022-04-01

How to you find the general solution of

\frac{d y}{d x} = x {\cos x}^{2}

delai59qk

Beginner2022-04-02Added 8 answers

First we notice that

d \frac{{\sin x}^{2}}{d x} = 2 x {\cos x}^{2}

x {\cos x}^{2} = \frac{1}{2} \cdot (d \frac{{\sin x}^{2}}{d x})

Hence the problem becomes

\frac{d y}{d x} = \frac{1}{2} \cdot \frac{d {\sin x}^{2}}{d x}

Integrate both sides with respect to 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

y = \frac{1}{2} \cdot d x = \frac{1}{2} \cdot \int (d \frac{{\sin x}^{2}}{d x}) d x

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

The general solution is

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

Nunnaxf18

Beginner2022-04-03Added 18 answers

\frac{d y}{d x} = x {\cos x}^{2}

d y = x {\cos x}^{2} d x

\int d y = \int x {\cos x}^{2} d x

Let

u = x^{2}

. Then

d u = 2 x d x and d x = \frac{d u}{2 x}

\int d y = \int x \cos u \frac{d u}{2 x}

\int d y = \int \frac{1}{2} \cos u d u

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

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

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?