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Finding y(t) from ODE (HELP NEEDED QUICK)

  • Thread starter andrey21
  • Start date
  • #1
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Find y(t) from the following first order ODE



Homework Equations



y' = 20SQRT10 x (1-e^(t/SQRT10))/(1+e(t/SQRT10))

The Attempt at a Solution


I know I must integrate to gain the solution but am having problems getting started please any help greatly appreciated!
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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It is a separation of variables type question. So try that. Then a substitution to make integration easier.
 
  • #3
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So far I have done the following:
dy/dt = 20SQRT10 x (1-e^(t/SQRT10))/(1+e(t/SQRT10))
dy . 1/20SQRT10 = (1-e^(t/SQRT10))/(1+e(t/SQRT10)) dt

I can then integrate the LHS to give,
y/20SQRT 10 WHich I think is correct?

How could I tackle the RHS? Tried numerous methods!!!
 
  • #4
rock.freak667
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[tex]u=e^{\frac{t}{\sqrt{10}}}[/tex]


Then split into partial fractions.
 
  • #5
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so if u=e^t/SQRt10 the RHS becomes:

∫ (1- u)÷(1+u) dt

From there you say split into partial fractions like the following ths is the part that is confusing me. How would you go about doing that?
 
  • #6
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Okay from integrating I have got:

t-te^(t/SQRT10)÷ e^(t/SQRT10) + 1

is this correct?
 
  • #7
rock.freak667
Homework Helper
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I think you forgot a 'u' in the denominator. You should get

[tex]\int \frac{1-u}{u(1+u)}du[/tex]

Just put

[tex]\frac{1-u}{u(1+u)} = \frac{A}{u}+\frac{B}{1+u}[/tex]

solve for A and B (the expression is true for all values of u)
 
  • #8
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Thanks using your method I have gotten,

A=1 and B = -2 which gives

1/(u) - 2/(1+u)

1/e^(t/SQRT10) - 2/(1+e^(t/SQRT10))

correct?
Giving me a final answer of:

Y/20SQRT(10) = 1/e^(t/SQRT10) - 2/(1+e^(t/SQRT10))
 
  • #9
rock.freak667
Homework Helper
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Thanks using your method I have gotten,

A=1 and B = -2 which gives

1/(u) - 2/(1+u)

1/e^(t/SQRT10) - 2/(1+e^(t/SQRT10))

correct?
Giving me a final answer of:

Y/20SQRT(10) = 1/e^(t/SQRT10) - 2/(1+e^(t/SQRT10))
But you do understand how there is an extra 'u' right and how partial fractions work?

Also, you are to integrate it, not just immediately replace u.
 
  • #10
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Okay Ive completly forgotton to integrate!

∫1/(u) - 2/(1+u) = ln(u) - 2ln(1+u)

Then subs in value for u:
ln(e^t/SQRT10) - 2ln(1+e^(t/SQRT10)

As ln and e cancel each other out this becomes:

t/SQRT10 - 2ln1 - 2ln(e^(t/SQRT10)

which becomes:

t/SQRT10 - 2(t/SQRT10)

is that correct?

To answer other question Im not clear where you get the extra u from!
 
  • #11
rock.freak667
Homework Helper
6,230
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Then subs in value for u:
ln(e^t/SQRT10) - 2ln(1+e^(t/SQRT10)

As ln and e cancel each other out this becomes:

t/SQRT10 - 2ln1 - 2ln(e^(t/SQRT10)
From here ln(a+b)≠ lna + lnb, you will need to leave it as is

To answer other question Im not clear where you get the extra u from!
Say you had u=ekx, then du=kekx dx, and if u=ekx, then wouldn't du=ku dx and thus dx= du/(ku) ?
 
  • #12
466
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Ah yes I see now where the extra u comes from so just to finish, the final answer I have taking everything into account is:

Y/20SQRT(10) = t/SQRT(10)-2ln(1+e^(t/SQRT10)

And from there I can simply solve to get equation in terms of Y as originally asked at the start.

Y = (t/SQRT(10)-2ln(1+e^(t/SQRT10) ) x 20SQRT(10)

This I know will simplify
 
  • #13
rock.freak667
Homework Helper
6,230
31
Ah yes I see now where the extra u comes from so just to finish, the final answer I have taking everything into account is:

Y/20SQRT(10) = t/SQRT(10)-2ln(1+e^(t/SQRT10)

And from there I can simply solve to get equation in terms of Y as originally asked at the start.

Y = (t/SQRT(10)-2ln(1+e^(t/SQRT10) ) x 20SQRT(10)

This I know will simplify
You've already found y(t) so all should be well now.
 
  • #14
466
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Thanks for help :)
 

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