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

Find y(t) from the following first order ODE



2. Relevant equations

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

3. 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!

 

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!!!

 

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?

 

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))

 

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!

 

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

 

Thanks for help :)