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amberjordan2006 Group Title

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  • one year ago
  • one year ago

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  1. Spacelimbus Group Title
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    \[y^3+\cos(t)y'=2+y \sin(t)\]is this the question?

    • one year ago
  2. Spacelimbus Group Title
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    does not look exact though without an integration factor.

    • one year ago
  3. Spacelimbus Group Title
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    \[y^3-y \sin(t)-2+\cos(t)y'=0\]

    • one year ago
  4. Spacelimbus Group Title
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    the y(0)=-1 is just an initial condition, the test for exactness is applied before that: \[M_y=N_x\]

    • one year ago
  5. Spacelimbus Group Title
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    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.

    • one year ago
  6. Spacelimbus Group Title
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    Usually you continue from this with an integration factor, but it doesn't say that in the given problem.

    • one year ago
  7. Spacelimbus Group Title
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    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.

    • one year ago
  8. Spacelimbus Group Title
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    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.

    • one year ago
  9. Spacelimbus Group Title
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    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: \[f_t=-y \sin (t) -2 \]

    • one year ago
  10. Spacelimbus Group Title
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    not quite, and \[f_y=(y^3+\cos(t))\]

    • one year ago
  11. Spacelimbus Group Title
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    without the y'

    • one year ago
  12. Spacelimbus Group Title
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    yes, what will you get from that?

    • one year ago
  13. Spacelimbus Group Title
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    You're integrating with respect to dy, so you should get: \[H(t,y)=\frac{ y^4 }{ 4 }+y \cos (t) +g(t)\]

    • one year ago
  14. Spacelimbus Group Title
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    yeh you got it almost.

    • one year ago
  15. Spacelimbus Group Title
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    very good, but also check for the cos(t) term, it's just a constant, so you add a y infront of it.

    • one year ago
  16. Spacelimbus Group Title
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    yes because your dimension/direction of integration is dy, so t is constant.

    • one year ago
  17. Spacelimbus Group Title
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    yes, how would you continue?

    • one year ago
  18. Spacelimbus Group Title
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    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)

    • one year ago
  19. Spacelimbus Group Title
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    yes, integrate that.

    • one year ago
  20. Spacelimbus Group Title
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    almost it would give you ycos(t)-2t+g(y)

    • one year ago
  21. Spacelimbus Group Title
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    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.

    • one year ago
  22. Spacelimbus Group Title
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    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: \[ H(t,y)=y \cos (t) -2t + \frac{ y^4 }{ 4 }+C\]

    • one year ago
  23. Spacelimbus Group Title
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    yes you still need to find C, do that with your initial values.

    • one year ago
  24. Spacelimbus Group Title
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    I believe H(t,y(0))=-2t+C=-1

    • one year ago
  25. Spacelimbus Group Title
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    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.

    • one year ago
  26. Spacelimbus Group Title
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    maybe you want to copy both answers to your problem sheet though, whichever your Professors prefer.

    • one year ago
  27. Spacelimbus Group Title
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    you're done

    • one year ago
  28. Spacelimbus Group Title
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    no there is only one constant of integration

    • one year ago
  29. Spacelimbus Group Title
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    you're welcome

    • one year ago
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