Question

(secx)dy/dx=e^(y+sinx), please help me solve the differential equation. Thanks :)

Answer 1

move secx to the other side

Answer 2

then moved dx to the other side, then integrate it. THAT"S ALL

Answer 3

\[\large \sec x\frac{dy}{dx}=e^{(y+\sin x)}\]

Recall a rule of exponents:\[\large e^{a+b}=e^a\cdot e^b\]

Answer 4

\[\large \sec x\frac{dy}{dx}=e^{(y+\sin x)} \qquad \rightarrow \qquad \large \sec x\frac{dy}{dx}=e^{y}\cdot e^{\sin x}\]So that would be our first step right?

Answer 5

Yes.

Answer 6

and then it would be \[e^y/secx \times e^{sinx}/secx\]

Answer 7

First order Seperable differential equation. You can think of it as "cross multiplying" Even though that's what it seems like, its not. but you could think of it as doing just that. You're getting dy w/ y's and dx with x's.

Answer 8

Yes. Next move the e^y to the side containing dy.

Answer 9

Woops the way you wrote it joker, it looks like your denominator is sec^2x. Careful. There should only be 1 secx on the bottom on the right side of our equation.

Answer 10

so \[\frac{ 1 }{ e^y }dy=e^{sinx}/secxdx\]?

Answer 11

\[\huge e^{-y}dy\qquad =\qquad \frac{e^{\sin x}}{\sec x}dx\]

Yes very good. We can simplify the 1/sec x. Remember what that will give us in terms of sines or cosines?

Answer 12

\[e^{-y}dy=e ^{sinx}times cosx\]

Answer 13

1/secx becomes cosx

Answer 14

\[\huge e^{-y}dy\qquad =\qquad (\cos x) e^{\sin x}dx\]

Ok good :) From here you can integrate.

Answer 15

do I use natural log?

Answer 16

You will use a `U substitution` to get through these integrals.

If your problem is asking for an `explicit` solution, then yes we'll have to use natural log after we integrate.

Answer 17

\[\frac{ 1 }{ \sec(x) } = \cos(x)\]

Answer 18

so then its u=sinx, du=cosx dx

Answer 19

\[e^{-y} dy=e^udu\]?

Answer 20

k looks good so far.

Answer 21

after taking the log \[-ydy=udu\]

Answer 22

No don't take the log right now. Integrate first.

Answer 23

after integrating \[-e^{-y}=e^u\]

Answer 24

\[\large -e^{-y}=e^u+C\]

Ok good.

Answer 25

Make sure to move that negative over before applying the natural log.

Answer 26

so then it becomes \[e^{-y}=-e^u ?\]

Answer 27

Don't forget your C, it's quite important in this problem.

Answer 28

\[\large e^{-y}=C-e^u\]

Answer 29

\[-y=C-u\]

Answer 30

so is the final answer \[y=sinx+c\] ?

Answer 31

Taking the natural log of both sides,\[\large \ln(e^{-y})=\ln(C-e^u)\]Giving us,\[\large -y=\ln(C-e^u) \qquad \rightarrow \qquad y=-\ln(C-e^{u})\]

Answer 32

Im not sure what happened to your natural log on the right side. It doesn't disappear like it does on the left.

Answer 33

oh, right, forgot the natural log on the other side. :) Thanks so much zepdrix! :)

Answer 34

np c: