# Second-Order separable Differential equations

1. Oct 11, 2013

### Woolyabyss

1. The problem statement, all variables and given/known data

Solve d2y/dt2 = dx/dt2, if x = 0 and dx/dt = 1 when t = 0

2. Relevant equations

3. The attempt at a solution

d2y = dx

I'm not exactly sure what to do here the fact that dt2 is under the denominator for both fractions is confusing memaybe its a typo? should it be d2y/dx2 = dx/dt?

2. Oct 11, 2013

### Staff: Mentor

That has to be a typo. dx/dt2 makes zero sense.

On a side note, try to make you posts clearer by at least indicating that some things are exponents. The simplest way is to use the caret or circumflex character (^), which is pretty much universally used for this purpose. For example, 3x2 and e^(rt).

A little nicer is to use the advanced menu (click Go Advanced below the input area. For exponents, click the X2 button. You can do subscripts with the X2 button.

For fancier stuff, you can use LaTeX to write things like $10x^2$ and even fancier stuff. Here's a link to a summary of how to do that: https://www.physicsforums.com/showthread.php?t=617567 - item 2 on the list.

3. Oct 11, 2013

### Woolyabyss

Alright thanks, If i were to assume they meant d^2y/dt^2 = dx/dt

would I be correct in saying I could integrate both sides and it would be dy/dt = x + c?

4. Oct 11, 2013

### Staff: Mentor

It might be better to figure out what the exact problem should be. Can you contact your instructor to get this clarified?

5. Oct 12, 2013

### Woolyabyss

No im afraid not I'll just leave ut for now. Thans anyway

6. Oct 12, 2013

### HallsofIvy

Staff Emeritus
The difficulty is that you have a single equation for two unknown functions, x and y. That is not sufficient. You need another equation.