Second order differential equations

[Doraamar]

Try to calculate the following equation from dx/dy = 1/y' :
1)d^2x/dy^2 = -y''/(y')^3

Hello, I am taking a calculus course in my university. I am not very good in it, so I apologize if my question is too 'dumb'.

Here is how I've been trying to solve this question:
I tried multiplying the first equation ( dx/dy = 1/y') by the integrating factor, and then integrate it, but for me it didn't work, I am sure I am doing something wrong. I also tried to find a number that satifies both equations, like a number N that the second derivative is equal to N^3, but that is also wrong too..
I feel like I keep hitting a wall when trying to solve it, so I will be thankful if anyone can help

 

[Deleted member 4993]

Try to calculate the following equation from dx/dy = 1/y' :
1)d^2x/dy^2 = -y''/(y')^3

Hello, I am taking a calculus course in my university. I am not very good in it, so I apologize if my question is too 'dumb'.

Here is how I've been trying to solve this question:
I tried multiplying the first equation ( dx/dy = 1/y') by the integrating factor, and then integrate it, but for me it didn't work, I am sure I am doing something wrong. I also tried to find a number that satisfies both equations, like a number N that the second derivative is equal to N^3, but that is also wrong too..
I feel like I keep hitting a wall when trying to solve it, so I will be thankful if anyone can help

Can you post a "photograph" of the original problem?

 

[Romsek]

Looks like it's just the quotient and chain rules for derivatives. Let's see.

[MATH]\dfrac{dx}{dy} = \dfrac{1}{y^\prime}[/MATH]
[MATH]\dfrac{d^2x}{dy^2} = \left(\dfrac{d}{dy} \dfrac{dx}{dy}\right)\dfrac{dx}{dy} = \\ \dfrac{0-1\cdot y^{\prime\prime}}{(y^\prime)^2}\cdot\dfrac{1}{y^\prime} = \\ -\dfrac{y^{\prime\prime}}{(y^\prime)^3} [/MATH]

 

[HallsofIvy]

Try to calculate the following equation from dx/dy = 1/y' :
1)d^2x/dy^2 = -y''/(y')^3

Hello, I am taking a calculus course in my university. I am not very good in it, so I apologize if my question is too 'dumb'.

Here is how I've been trying to solve this question:
I tried multiplying the first equation ( dx/dy = 1/y') by the integrating factor, and then integrate it, but for me it didn't work, I am sure I am doing something wrong.

What does y' indicate here? If it is the derivative, \(\displaystyle \frac{dy}{dx}\),then \(\displaystyle \frac{dx}{dy}= \frac{1}{y'}= \frac{1}{\frac{dy}{dx}}\) is true for any x and y such that both derivatives exist.

I also tried to find a number that satifies both equations, like a number N that the second derivative is equal to N^3, but that is also wrong too..
I feel like I keep hitting a wall when trying to solve it, so I will be thankful if anyone can help

??? The derivative of any number (constant function) is 0. How could you possibly say that "the second derivative is N^3"?

 

[Steven G]

You keep referring about the work you have done. Why not share that work so we can see exactly what is going on. Just for the record, the sharing of your work is part of the posting guideline for this forum. It makes things go much faster if we get to see your work.