Recent content by gerechte23

can someone help me to find an function Y of which Y'(0)=1/2, Y''(0)=1=2/2.. the nth derivative verifies Y''''''''(0)=n/2 thanks for your help

Here is the equation it gave me. I resolved it, i didn't expect to come to the internet so i didn't come with the papers, so is it right? \int\sqrt{\frac{y}{by-c}} dy = \int\sqrt{2} dx. b and c are real numbers!

hi buddy, Thanks for your help. Actually, it's the very same method a man here has proposed to me. And i ended up having one equation where i have my Y on one side, and the variable x on the other. When i kept integrating it gives me something with logarithm and square root with Y. i wanted to...

Thanks for the source. I'll make my best to solve it. But i would like to have a person to help me with it though. I barely can use the internet. I am in Haiti. Can u try helping me with it please!

Hi I've been striving to find the solution of this differential equation, but can't find: Y''*Y^2=C c being a real number. Please give me a hand with it. Thanks in advance!

hi guys, maybe the way I've started resolving the equation is too difficult. Can someone help me with this please: Y''*Y^2=C, c is a real number. Please, help me with tis, i am killing myself to find that solution but failing all the time. Thanks guys for helping again* with it.

another question about that log in the result, can it be ln? (Neperian logarithm?)

hi again, yes. I want to calculate it, but it's not a homeworh though. I've proposed myself to find it. Cause with it, i'd be able to calculate a certain acceleration depending on time. Don't you know a way that could help me solve it. If yes, please give a hand

Thanks for all guys. Now my only problem is to calculate y as a function of t [Y(t)]. It seems quite complicated for me though. Thanks again

Excuse me, but can you help me in this. I just can't locate where to click so that i start a new thread in here. Please help

\int\sqrt{Y}/\int\sqrt{DY-C} = \int\sqrt{2}dt . When i resolve this equation using variable change i found the first part like: -\sqrt{[(DY-C)/D]} which seems not correct when i derivate it, it doesn't give me the result i was expecting. But when i use trigonometry variable changes, it give...

friend, i was working on the result and then i realized this :(1/2)(Y')^2= -C/Y+ D= (DY- C)/Y or Y'= 2(DY- C)/Y . The problem in this is Y'=[2(DY- C)/Y]^1/2. So, a little error came in. But I've tried solving it, but no. I am not good at all in differential equations. Please give me a hand again...

Hello, I am calculating a very important thing. But to find it, I should be able to resolve this equation. Y^2xY’’=C (C: Real number). Please help me solve it. Thanks in advance! (YY)xY''=C (C:Real nnumber)