DE:y' + 2ty = 5t

  • Thread starter luitzen
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  • #1
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Homework Statement


Solve DE:y' + 2ty = 5t

The Attempt at a Solution


For yh: y' + 2ty = 0
y' = -2ty
Thus int(y-1,y) = int(-2t,t)
ln(y) = -t2 + C1
y = e-t2 + C1 = eC1e-t2 = Ce-t2

To find yp there are two ways to do it. With an integrating factor exp(-A(t)) where A'(t) = 2t or by saying yp = v(t)yh(t). Both ways will result in int(e-t2,t) which is unsolvable.

So I checked up the answer which read 5/2 + Cet2, so yp(t)=5/2. So I checked it by entering 5/2 in the DE and of course it was right.

Now my question is, how would you arrive at that answer? When would you make the assumption yp' = 0 and check if it is right? Isn't that too trivial? And when you would find an answer for yp' = 0, would you stop searching for another answer?
 
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Answers and Replies

  • #2
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Hi luitzen! :smile:

Can't you just argue that

[tex]y^\prime+2ty=5t[/tex]

is equivalent with

[tex]y^\prime=t(5-2y)[/tex]

and thus

[tex]\frac{y^\prime}{5-2y}=t[/tex]
 
  • #3
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Yes, of course. The question was: Check if the following equations are separable equations and if yes, give the solution.

So, if you have a separable equation, you don't need to find yh and yp, am I right?

If that's not the case, you try to find yh and after that yp, right?

Thank you very much:)
 
  • #4
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Yes, of course. The question was: Check if the following equations are separable equations and if yes, give the solution.

So, if you have a separable equation, you don't need to find yh and yp, am I right?

Right! Seperable equations are easy!

If that's not the case, you try to find yh and after that yp, right?

Indeed, if the equation is not seperable, then you'll need to do other tricks. For example, working with yh and yp...
 
  • #5
LCKurtz
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In my opinion, the best way to have worked that problem is to observe that it is a linear first order so you can multiply by [itex]e^{t^2}[/itex] making an exact derivative.
 
  • #6
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How would you integrate et2?
 
  • #7
vela
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You shouldn't have to. How are you coming up with that integral?
 
  • #8
LCKurtz
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How would you integrate et2?

You wouldn't:
[tex]e^{t^2}y'+2te^{t^2}y=5te^{t^2}[/tex]
[tex](ye^{t^2})' =5te^{t^2}[/tex]
 
  • #9
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Ok this is what I did:
[itex]2ty-5t+y'=0[/itex]

[itex]e^{t^{2}}\left(2ty-5t\right)+e^{t^{2}}y'=0[/itex]

[itex]\dfrac{\partial F\left(y,t\right)}{\partial t}=e^{t^{2}}\left(2ty-5t\right)[/itex]

[itex]\dfrac{\partial F\left(y,t\right)}{\partial y}=e^{t^{2}}[/itex]
 
  • #10
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You wouldn't:
[tex]e^{t^2}y'+2te^{t^2}y=5te^{t^2}[/tex]
[tex](ye^{t^2})' =5te^{t^2}[/tex]
If I follow that route, I'll get this:

[itex]\left(ye^{t^{2}}\right)'=5te^{t^{2}}[/itex]

[itex]ye^{t^{2}}=\int5te^{t^{2}}dt[/itex]

[itex]\int5te^{t^{2}}dt=5t\int e^{t^{2}}dt-5\int\int e^{t^{2}}dtdt[/itex]

[itex]y=5e^{t^{-2}}\left(t\int e^{t^{2}}dt-\int\int e^{t^{2}}dtdt\right)[/itex]
 
  • #11
HallsofIvy
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No, no, no!

Let [itex]u= t^2[/itex]. Then du= 2t dt, tdt= du/2.
 
  • #12
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lol, I think I use integration by parts too quickly.

It's not too hard, but I still don't get it why it's easier though.
 

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