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

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!

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rock.freak667
Homework Helper
It is a separation of variables type question. So try that. Then a substitution to make integration easier.

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!!!

rock.freak667
Homework Helper
$$u=e^{\frac{t}{\sqrt{10}}}$$

Then split into partial fractions.

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?

Okay from integrating I have got:

t-te^(t/SQRT10)÷ e^(t/SQRT10) + 1

is this correct?

rock.freak667
Homework Helper
I think you forgot a 'u' in the denominator. You should get

$$\int \frac{1-u}{u(1+u)}du$$

Just put

$$\frac{1-u}{u(1+u)} = \frac{A}{u}+\frac{B}{1+u}$$

solve for A and B (the expression is true for all values of u)

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))

rock.freak667
Homework Helper
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.

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!

rock.freak667
Homework Helper
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) ?

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

rock.freak667
Homework Helper
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.

Thanks for help :)