Question

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

Answer 1

What do you want to do with it?

Answer 2

Solve for y using integration

Answer 3

I know I'd get 1+y^2/y=cosx, but the integration by substitution at that point gets a bit confusing

Answer 4

\[\frac{ dy }{ dx } = \frac{ y \cos(x) }{ 1 + y^2 }\]
\[\int\limits\frac{ (1 + y^2)dy }{ y } = \int\limits \cos(x)dx\]

Answer 5

Well, this is a differential equation because, we can see that a derivative appears there, so:
\[\frac{ dy }{ dx }=\frac{ ycosx }{ 1+y^2 }\]
so, leaving it to one side:
\[\frac{ (1+y^2)dy }{ y }=cosxdx\]
And now, it's a matter of integrating on both sides:
\[\int\limits \frac{ (1+y^2)dy }{ y }=\int\limits cosx dx\]
I will take the first steps on the left:
\[\int\limits \frac{ (1+y^2)dy }{ y }\]
since the grade of the numerator is higher than the denominator, we can divide both polynomials:
\[\int\limits \frac{ y^2 }{ y }+\frac{ 1 }{ y }dy\]
\[\int\limits\limits \frac{ y^2 }{ y }+\frac{ 1 }{ y }dy = \int\limits ydy + \int\limits \frac{ 1 }{ y }dy\]
\[\int\limits \frac{ (1+y^2)dy }{ y }=\frac{ y^2 }{ 2 }+ Ln(y) +c\]

Answer 6

\[\int\limits(\frac{ 1 }{ y })dy + \int\limits y dy = \sin(x) \]

Answer 7

\[\frac{ y^2 + Ln ( y^2) }{ 2 }=sinx\]

Answer 8

Okay, I see how that works Owlcoffee, thank you. When trying to solve for y what do I do when there are two ys?

Answer 9

I'm not sure... I can't seem to be able to isolate the y...

Answer 10

It's okay, I cant either

Answer 11

Are you sure the problem is like that?

Answer 12

Oh, you just have to plug in the condition given with the problem.

Answer 13

I see, so solving wasn't needed.

Answer 14

No, it was you just plug in the conditions after you get it to that point and I saw on Khan Academy that that is an infamous differential equation and you cant get any further after solving for C

Answer 15

I'll have that in mind, thank you.

Answer 16

Sure, thank you.