## Denebel 3 years ago use separation of variables to solve the initial value problem. dy/dx=(cos x)e^(y+sinx) and y=0 when x=0

1. myininaya

$e^{-y} dy =\cos(x)e^{\sin(x)} dx$

2. myininaya

integrate both sides

3. Denebel

Lol.. I don't know how to move "the x with dx" and "y with dy".. :/ can you show me that part?

4. myininaya

i multiplied dx on both sides

5. myininaya

$e^{y+\sin(x)}=e^{y} e^{\sin(x)}$

6. myininaya

divide both sides by e^y

7. myininaya

1/e^{y}=e^{-y}

8. Denebel

Oh ok. Thanks.

9. Denebel

How do I integrate cosxe^(sin x) dx?

10. myininaya

ley u=sin(x) => du=cos(x) dx

11. Denebel

? What about the e? Do I put e to the u power?

12. myininaya

$\int\limits_{}^{}e^u du=e^u+C=e^{\sin(x)}+C$

13. myininaya

i replace sin(x) with u and cos(x) dx with du

14. Denebel

Oh? But don't I integrate |dw:1327549907891:dw|? What happens with the cos...

15. myininaya

cos(x) dx=du i replaced cos(x) dx with du i did integrate

16. myininaya

|dw:1327550585485:dw|

17. myininaya

|dw:1327550603285:dw|

18. Denebel

Ohh I see. Ok, what do I do now? Do I plug in the given x, y values to find C?

19. myininaya

have you integrated the other side yet?

20. Denebel

I have now: -e^(-y) = e^(sinx) + C

21. myininaya

ok so enter in (0,0) and solve for C

22. Denebel

I solved that C = 0, So does the equation become -e^(-y) = e^(sinx) ?

23. myininaya

-1-1=-2

24. myininaya

-1=1+C -1-1=C -2=C

25. Denebel

Oh I forgot about the negative on the left.

26. Denebel

Then -e^(-y) = e^(sinx) - 2 ?

27. myininaya

yes

28. Denebel

Thanks!