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SpacelimbusBest ResponseYou've already chosen the best response.0
\[y^3+\cos(t)y'=2+y \sin(t)\]is this the question?
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
does not look exact though without an integration factor.
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
\[y^3y \sin(t)2+\cos(t)y'=0\]
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
the y(0)=1 is just an initial condition, the test for exactness is applied before that: \[M_y=N_x\]
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
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.
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
Usually you continue from this with an integration factor, but it doesn't say that in the given problem.
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
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.
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
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.
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
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: \[f_t=y \sin (t) 2 \]
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
not quite, and \[f_y=(y^3+\cos(t))\]
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
yes, what will you get from that?
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
You're integrating with respect to dy, so you should get: \[H(t,y)=\frac{ y^4 }{ 4 }+y \cos (t) +g(t)\]
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
yeh you got it almost.
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
very good, but also check for the cos(t) term, it's just a constant, so you add a y infront of it.
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
yes because your dimension/direction of integration is dy, so t is constant.
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
yes, how would you continue?
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
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)
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
yes, integrate that.
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
almost it would give you ycos(t)2t+g(y)
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
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.
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
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: \[ H(t,y)=y \cos (t) 2t + \frac{ y^4 }{ 4 }+C\]
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
yes you still need to find C, do that with your initial values.
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
I believe H(t,y(0))=2t+C=1
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
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.
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
maybe you want to copy both answers to your problem sheet though, whichever your Professors prefer.
 11 months ago

SpacelimbusBest ResponseYou've already chosen the best response.0
no there is only one constant of integration
 11 months ago
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