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

• andrey21
In summary, the person is trying to solve a separation of variables equation but is having difficulty. They have done some integration and found that t/SQRT10 - 2ln(1+e^(t/SQRT10))=-2. They have also substituted in u=ekx to get y/20SQRT10=-1/e^(t/SQRT10) and solved for Y.

#### andrey21

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!

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!

$$u=e^{\frac{t}{\sqrt{10}}}$$

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?

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))

Jamiey1988 said:
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 I've completely 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 I am not clear where you get the extra u from!

Jamiey1988 said:
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

Jamiey1988 said:
To answer other question I am 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

Jamiey1988 said:
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 :)

## 1. How do I find the solution to an ODE (ordinary differential equation)?

There are various methods for solving ODEs, including separation of variables, substitution, and use of integrating factors. It is important to first identify the type of ODE and choose an appropriate method for solving it.

## 2. Can I use a computer program to solve ODEs?

Yes, there are many software programs and online resources available for solving ODEs. Some popular options include MATLAB, Mathematica, and Wolfram Alpha.

## 3. What is the difference between an explicit and implicit solution for an ODE?

An explicit solution expresses the dependent variable y in terms of the independent variable x, while an implicit solution does not explicitly solve for y. Implicit solutions are often used for more complex ODEs that cannot be easily solved explicitly.

## 4. How do I know if my solution to an ODE is correct?

One way to check the accuracy of your solution is to plug it back into the original ODE and see if it satisfies the equation. Additionally, you can compare your solution to other known solutions or use numerical methods to approximate the solution.

## 5. Are there any common mistakes to avoid when solving ODEs?

Some common mistakes include incorrect application of a method, incorrect use of initial conditions, and algebraic errors. It is important to double check your work and seek help if you are unsure about a step in the process.