anonymous 3 years ago .

1. anonymous

$y^3+\cos(t)y'=2+y \sin(t)$is this the question?

2. anonymous

does not look exact though without an integration factor.

3. anonymous

$y^3-y \sin(t)-2+\cos(t)y'=0$

4. anonymous

the y(0)=-1 is just an initial condition, the test for exactness is applied before that: $M_y=N_x$

5. anonymous

Show that the differential equation (y^3)+cos(t)y'=2+ysin(t), y(0)=-1 is exact and solve the i.v.p - It is not exact to my understanding. So I (Or maybe the problem) fail at the first place.

6. anonymous

Usually you continue from this with an integration factor, but it doesn't say that in the given problem.

7. anonymous

I can have a look into it, but I believe that for this problem it will turn out to become rather hairy. Are you really sure that you copied the problem correct? I have a strange feeling about this. Not that I say that every problem is always to be correct, but the phrasing would make absolutely zero sense that way. On an exam for instance, you could show that the given DE was not exact and then drop it, because it's a contradiction to the given statement.

8. anonymous

See, you forgot about some important paranthesis in your opening post, or I didn't quite understand them. $(y^3+\cos(t))y'=2+y \sin(t)$ not that is the same as: $-y \sin(t) -2 + (y^3+ \cos(t))y'=0$ and now you can indeed apply the test and see that this statement holds true: $M_y=N_t$ and thus it is exact.

9. anonymous

They should be partial differentials, you're dealing with multivariable calculus (Differential Equations) here, the above emerges from the multivariable chain rule. Since it is exact you should: $f_t=-y \sin (t) -2$

10. anonymous

not quite, and $f_y=(y^3+\cos(t))$

11. anonymous

without the y'

12. anonymous

yes, what will you get from that?

13. anonymous

You're integrating with respect to dy, so you should get: $H(t,y)=\frac{ y^4 }{ 4 }+y \cos (t) +g(t)$

14. anonymous

yeh you got it almost.

15. anonymous

very good, but also check for the cos(t) term, it's just a constant, so you add a y infront of it.

16. anonymous

yes because your dimension/direction of integration is dy, so t is constant.

17. anonymous

yes, how would you continue?

18. anonymous

Did you mean to integrate the above? Or differentiate again to figure out what g(t) is? There is more than one way to do that. - 1) You have H(t,y) now, but you're missing g(t), so you can partial differentiate H(t,y) with respect to t, you can set that equal to the already given given H_t - 2) You can integrate the second given partial DE and then match and compare coefficients to figure out g(x)

19. anonymous

yes, integrate that.

20. anonymous

almost it would give you ycos(t)-2t+g(y)

21. anonymous

So now you have two equations which you can match: $H(y,t)=\frac{ y^4 }{ 4 }+y \cos(t) + g(t)$ and $H(y,t)=y \cos(t) -2t + g(y)$ So you can try to figure out maybe from there how to continue.

22. anonymous

first you need to figure out your g(t) and g(y), g(t) is obviously a function that only depends on t, so you can see that g(t)=-2t and g(y) is a function that only depends on y, so g(y)=y^4/4 So your solution is: $H(t,y)=y \cos (t) -2t + \frac{ y^4 }{ 4 }+C$

23. anonymous

yes you still need to find C, do that with your initial values.

24. anonymous

I believe H(t,y(0))=-2t+C=-1

25. anonymous

I think it's better to keep t as a function itself, unfortunately the above function cannot be expressed in an explicit form y(t)=... So I would just let the t remain there, because they aren't giving any additional infos about it.

26. anonymous

27. anonymous

you're done

28. anonymous

no there is only one constant of integration

29. anonymous

you're welcome