# Solve x'' = cox(x)

• peacex

#### peacex

im new to 2nd order differential eqns and I am getting myself all confused.
ive tried so hard but i just can't solve it.

x'' = cos(x)

Try using $$\ddot{x} = \dot{x}\frac{d\dot{x}}{dx}$$

That will allow you to turn it into a first order equation at the very least.

thankyou! i didnt think of that!

Try using $$\ddot{x} = \dot{x}\frac{d\dot{x}}{dx}$$

That will allow you to turn it into a first order equation at the very least.

I don't think that's correct.
You should have written

$$\ddot{x}=\frac{1}{\dot{x}}\frac{d}{dt}\left(\frac{\dot{x}^{2}}{2}\right)$$

No, Dextercioby, the first response was correct. If you let u= dx/dt, then du/dt= d2x/dt2. But we can apply the chain rule to the left side: du/dt= (du/dx)(dx/dt)= u du/dx since u= dx/dt. d2u/dt2= u du/dx.

This method is called "quadrature" (because the integral involves u2) and can be used to reduce the order of a differential equation when the independent variable, t, does not appear explicitely in the equation.

x" = cosx

its x-missing

v = x'

v' = x" = v dv/dx

v dv/dx = cosx

vdv = cosx dx

v^2 = 2 sinx

(x')^2 = 2 sinx

x' = ( 2 sinx )^1/2

dx / (sinx )^1/2 = 2^1/2 dy

Q = 2^1/2 y + C

such that Q = http://integrals.wolfram.com/index.jsp

Last edited:
You cannot neglect the constant of integration.

if

$$y^{\prime \prime} = \cos{y}$$

multiply by $$y^{\prime}$$

gives you

$$y^{\prime} y^{\prime \prime} = y^{\prime} \cos{y}$$

integrate, and you get

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

Note the constant of integration which must not be ignored.

Therefore, you have

$$y^{\prime} = \sqrt{2 \sin{y} + C}$$

which isn't easily soluble -- the only way to get an answer is to use elliptic integrals -- have a look at this Wolfram page.

No, Dextercioby, the first response was correct.

It can't be, since $$\frac{d\dot{x}}{dx}=0$$.

Where in the world did you get that idea?

If x(t)= et, then x'= et= x. In other words,
$$\frac{dx'}{dx}= 1$$
not 0.