Question

Diff Eq dy/dx= (ycos(x))/(1+y^2) initial condition is y(0)=1 work below

Answer 1

divided by y and multiply both sides by y^2 and got y^2/y dy = cos(x)dx integrate that and got y^2/2 = cos(x)+c and then got y= +/- sqrt (2cos(x)+2c) and am not sure where to go now

Answer 2

separation of variables again..
\[\frac{1+y^2}{y}\; dy = \cos(x)\; dx\]
\[(\frac{1}{y} + y)\; dy = \cos(x) \;dx\]
integrate both sides

Answer 3

You have to divide by the whole factor (1+y^2), can't break up across addition or subtraction.

Answer 4

Oh duh so that would give me +/- e^(sqrt(2cos(x)+2(c))

Answer 5

\[\ln(y) + \frac{1}{2}y^2 = -\sin(x) + c\]
When x = 0 y = 1, so plug in these numbers to solve for c.
\[\ln(1) + \frac{1}{2}(1)^2 = -\sin(0) + c\]
\[0 + \frac{1}{2} = 0 + c\]
\[\frac{1}{2} = c\]

Answer 6

then solve for c = y(0)=1?

Answer 7

I think you forgot to integrate cosine :)

Answer 8

Yea that too.

Answer 9

When should I satisfy initial conditions?

Answer 10

Once I get it into y= blah or do I try to satisfy for c when it is (function)y = function(x)

Answer 11

In this case, you can't solve specifically for y, so just plug in the initial condition. If you can solve for y, then it really doesn't matter when you solve for c. I usually do before I solve for y.

Answer 12

Thanks alot.