- #1

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## 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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- Thread starter andrey21
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- #1

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

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

- #2

rock.freak667

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- #3

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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.

Then split into partial fractions.

- #5

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∫ (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?

t-te^(t/SQRT10)÷ e^(t/SQRT10) + 1

is this correct?

- #7

rock.freak667

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[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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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

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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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∫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

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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=e

- #12

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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

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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

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Thanks for help :)

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