Question

Solve the initial value problem.

\[(x^2-\sin^2(y))dx+xsin(2y)dy = 0, y(1) = \frac{ \pi }{ 3 }\]

Find x.

Answer 1

First off, can you show that this is exact or not?

Answer 2

The equation is not exact.

Answer 3

Alright, so do you know how to make it exact with an integrating factor?

Answer 4

\[M _{y} = -2sinycosy\]
\[N_{x} = 2sinycosy\]
These two are not exact so you can find \[\frac{ M_{y} - N_{x} }{ N } = \frac{ -2 }{ x }\]

Answer 5

Integrating factor is \[x ^{-2}\]

Answer 6

Alright, cool. So let's keep this ball rolling, what's next?

Answer 7

Multiplying by the integrating factor gives the equation
\[(1-x^{-2}\sin^2y)dx + (2x^{-1}sinycosy)dy = 0\]
The equations are now exact.

Answer 8

Alright, so what's next?

Answer 9

The general solution should be \[\psi(x,y) = \int\limits_{}^{}(1-x^{-2}\sin^2y)dx = x+x^{-1}\sin^2y + c(y)\]
Therefore:
\[\psi _{y}' = 2x^{-1}sinycosy+c'(y) = N = 2x^{-1}sinycosy\]
Meaning c'(y) = 0 so
\[c(y) = c\]
Which should give the general solution
\[\psi(x,y) = x+x^{-1}\sin^2y=c\]

Answer 10

Using \[y(1) = \frac{ \pi }{ 3 }\]
\[1 + 1^{-1}\sin^2(\frac{ \pi }{ 3 }) = c\]
Therefore c = 7/4
Which means
\[x + x^{-1}\sin^2y = \frac{ 7 }{ 4 }\]
Is this correct?

Answer 11

Looks correct to me.

Answer 12

How do I solve x as a function of y using that solution?

Answer 13

just isolate y
sin^2y = -x^2 +(7/4)x

Answer 14

http://www.wolframalpha.com/input/?i=y%27%3D%28+sin%28y%29^2+-+x^2%29%2F%28x+sin%282y%29%29%2C++y%281%29%3DPi%2F3

Answer 15

take square root both sides and arcsin both sides to get y respect to x

Answer 16

The question is looking for x with respect to y, not y respect to x.

Answer 17

If so, you have x^2 -7/4 x + sin^2 y =0 right?
solve as a quadratic problem with a =1, b = -7/4 and c = sin^2 y
you may have 2 roots x1, and x2 consider the condition to get the right answer, you may have both roots . Got me?

Answer 18

@Loser66 is right, here are the roots
\[
x= \frac{1}{8} \left(7-\sqrt{49-64
\sin ^2(y)}\right)\\x= \frac{1}{8}
\left(7+\sqrt{49-64 \sin
^2(y)}\right)
\]