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# Solve this ODE? watch

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1. Does anyone have any idea how to solve this equation

d2y/dx2 = K / y2

where K is a constant?

Or indeed, does anyone know why it is hard to solve/whether I should do it numerically....
2. maybe:

=> dx^2/d^2y = y^2/k

int[d^2x] = (1/k) int[dy^2]

x^2/2 = (1/k)y^4/12 +c
3. actually might be

x^2/2 = (1/k)y^4/12 + ky + c

or even:

x^2/2 + nx = (1/k)y^4/12 + ky + c
4. Good try. That was one particular erroneous route I hadn't considered!

However, 1/(d^2y/dx^2) does not equal (d^2x/dy^2). That only works for dy/dx = 1/(dx/dy).

I've just put it into Maple and got a horrid answer out. I'm actually trying to find the equation of motion for a space vehicle powered by a solar sail, which x is the distance from the sun:

d2x/dt2 = K/x2

So I did:

v*dv/dx = K/x2

v2/2 = K(1/R - 1/x2)

where R is the radius of the sun (that's the closest u can start your solar sail vehicle!)

So that's the furthest I got!
5. Good try. That was one particular erroneous route I hadn't considered!

However, 1/(d^2y/dx^2) does not equal (d^2x/dy^2). That only works for dy/dx = 1/(dx/dy).

I've just put it into Maple and got a horrid answer out. I'm actually trying to find the equation of motion for a space vehicle powered by a solar sail, which x is the distance from the sun:

d2x/dt2 = K/x2

So I did:

v*dv/dx = K/x2

v2/2 = K(1/R - 1/x2)

where R is the radius of the sun (that's the closest u can start your solar sail vehicle!)

So that's the furthest I got
Ah well. What kind of level stuff is this?
6. Well, I'm a 2nd year physics undergrad. But the maths is probably A-level. I think it might be a non-linear equation, and I have no idea how to solve them, or approximately solve them.

I just wondered if any bright sparks had some ideas.
7. We never did these type of problems at A-level, but this is the sort of thing I've been doing in my first year of undergrad maths.

Write p = dy/dx

d2y/dx2 = d/dx(dy/dx) = dp/dx

dp/dy = (dp/dx)*(dx/dy) , by the chain rule
dp/dy = (d2y/dx2)(1/p)

=> d2y/dx2 = p(dp/dy)

=> p(dp/dy) = K/y2

INT (p)dp = INT (K/y2)dy

p2/2 = -K/y + c

p = [2(c - K/y)]1/2
=> dy/dx = [2c - (2K)/y]1/2

INT [2c - (2K)/y]-1/2 dy = INT dx

(-1/K)[2c - (2K)/y]1/2 = x + c'

And rearrange as necessary.
8. (Original post by ssmoose)
maybe:

=> dx^2/d^2y = y^2/k

int[d^2x] = (1/k) int[dy^2]

x^2/2 = (1/k)y^4/12 +c
There are several fundemental errors in your thinking:

1) dy/dx = 1/(dx/dy) !=> d2y/dx2 = 1/(d2x/dy2)

2) What do you assume "∫d2x" and "∫dy2" mean?

Whatever you think it means, it's pretty far from the truth! For example, try doing it on the equation d2x/dy2 = 1 (which has a fairly obvious solution). This is just an exercise in using the usual A-Level trick:

d2x/dt2 = v dv/dx

Where v = dx/dt.

9. (Original post by Wrangler)
This is just an exercise in using the usual A-Level trick:

d2x/dt2 = v dv/dx

Where v = dx/dt.

So how do you get x in terms of t for what i did?
10. v dv/dx = K/x2

vdv = Kdx/x2

v2/2 = -K/x + C

dx/dt = sqrt[ 2C - 2K/x ] = sqrt[2Cx-2K]/sqrt[x]

sqrt[x]dx/sqrt[Cx-K] = sqrt[2]dt

Let Cx = Kcosh2y
dx = (2K/C)coshy.sinhy.dy
sqrt[x] = sqrt[K/C]coshy
sqrt[Cx-K] = sqrt[K]sqrt[cosh2y-1] = sqrt[K]sinhy

Gives an integral of cosh2y dy = dt (with the appropriate multiplicative constants).
11. its further maths i think its differential equations
12. Awesome. Good work, AlphaNumeric.

I think I might have just rearranged something wrong to being with which is why Maple had a nightmare, because you get to a pretty standard integral, so there's no reason why it shouldn't have solved it. Ah well.
13. (Original post by Worzo)
So how do you get x in terms of t for what i did?
Magic, of course.

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