A cry for help... With Differential equations...

[Craig-Anthony]

Greetings fellow nerds, it is clearly evident that we come in all shapes sizes and from all the spectra of nerdy disciplines, I however appear to be required to master one that I am not at all adept at... Calculus x.x

 

I've been working on a three problem long assignment all day and haven't a clue what I'm supposed to be doing, if you can help me out in anyway it would be MASSIVELY appreciated~

 

This is the one I'm having the most problems with... I think if I could understand this one the rest wouldn't be so tough. It's an Initial value problem

Y(0)=1

 

dy/dx= (ycos(x))/(1+y^2)

 

Thank you again to anyone that can help me out!

[friendoftheood]

What are you trying to find? Are you trying to find the "initial value"? I don't know what that means, but if you explained it to me I could try to help. I love Calculus.

[Craig-Anthony]

I'm supposed to work out what equation makes the statement true. So some function of Y (or X) that makes the equation
dy/dx= (ycos(x))/(1+y^2)

True, such that when x is 0 y=1

Other then that I don't understand much myself

Ps. We should all be friends to the Ood, they're like adorable mindflayers.

[friendoftheood]

I have no idea. What Calc are you taking btw?

[Craig-Anthony]

Er, Calculus 2 at the Cegep level. It's part of the Quebec educational system I don't know what it would be anywhere else? probably first year of College or grade 13?

[friendoftheood]

I don't know if you've figured this problem out already, or if it's too late anyway, but I think I get the main idea. You have to take the integral of dy/dx to get the indefinite y function, and then using the fact that y(0)= 1 you know what your c is. (c being the constant that you have to add to the y function.) What I don't know how to do, though is take the integral of that function here. I've only done very simple integration so far.

[Craig-Anthony]

I did manage to work it out, the process is to isolate the equations so that you have some function of Y on one side and a function of X on the other.

Then you move dx to the X side giving

f(y) dy = f(x) dx

then you can integrate

F(y)=F(x) +C

And work it out just like you said. Thank you for all your help, very kind of you!

[friendoftheood]

It's no problem, it's actually fun for me. I love calculus. Math is kinda my thing.