Second order differential equations

Doraamar

New member
Joined
Jan 10, 2021
Messages
1
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 :)
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
23,400
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

Senior Member
Joined
Nov 16, 2013
Messages
1,290
Looks like it's just the quotient and chain rules for derivatives. Let's see.

\(\displaystyle \dfrac{dx}{dy} = \dfrac{1}{y^\prime}\)

\(\displaystyle \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}

\)
 
Top