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\[y^3+\cos(t)y'=2+y \sin(t)\]is this the question?
 one year ago

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does not look exact though without an integration factor.
 one year ago

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\[y^3y \sin(t)2+\cos(t)y'=0\]
 one year ago

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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

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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

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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

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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

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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

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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

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not quite, and \[f_y=(y^3+\cos(t))\]
 one year ago

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without the y'
 one year ago

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yes, what will you get from that?
 one year ago

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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

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yeh you got it almost.
 one year ago

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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

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

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yes, how would you continue?
 one year ago

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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

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yes, integrate that.
 one year ago

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almost it would give you ycos(t)2t+g(y)
 one year ago

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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

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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

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

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I believe H(t,y(0))=2t+C=1
 one year ago

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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

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maybe you want to copy both answers to your problem sheet though, whichever your Professors prefer.
 one year ago

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you're done
 one year ago

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no there is only one constant of integration
 one year ago

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you're welcome
 one year ago
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