- #1

- 4

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ive tried so hard but i just cant solve it.

can someone please help me with tips to solve/solutions?

x'' = cos(x)

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- Thread starter peacex
- Start date

- #1

- 4

- 0

ive tried so hard but i just cant solve it.

can someone please help me with tips to solve/solutions?

x'' = cos(x)

- #2

- 289

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

- #3

- 4

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thankyou!! i didnt think of that!

- #4

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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

[tex] \ddot{x}=\frac{1}{\dot{x}}\frac{d}{dt}\left(\frac{\dot{x}^{2}}{2}\right) [/tex]

- #5

HallsofIvy

Science Advisor

Homework Helper

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This method is called "quadrature" (because the integral involves u

- #6

- 9

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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

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:

- #7

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if

[tex]y^{\prime \prime} = \cos{y}[/tex]

multiply by [tex]y^{\prime}[/tex]

gives you

[tex]y^{\prime} y^{\prime \prime} = y^{\prime} \cos{y}[/tex]

integrate, and you get

[tex] \frac{1}{2}y^{\prime 2} = \sin{y} + C[/tex]

Note the constant of integration which must not be ignored.

Therefore, you have

[tex]y^{\prime} = \sqrt{2 \sin{y} + C} [/tex]

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

- #8

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No, Dextercioby, the first response was correct.

It can't be, since [tex] \frac{d\dot{x}}{dx}=0 [/tex].

- #9

HallsofIvy

Science Advisor

Homework Helper

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If x(t)= e

[tex]\frac{dx'}{dx}= 1[/tex]

not 0.

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