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# ODE question (integrating factor) Watch

1. I have:

that would go to:

Right? Not sure what to do next

Initial conditions are y = 0 at t = 0 and the constant is meant to equal

The answer is meant to be:

How did they get to this result?
2. (Original post by G A B R I E L)
I have:

that would go to:

Right? Not sure what to do next

Initial conditions are y = 0 at t = 0 and the constant is meant to equal

The answer is meant to be:

How did they get to this result?
I'm not sure what you mean by 'that would go to', as I have no idea what you've done there. The integrating factor for this is .
Now multiply through your differential equation by p(x) and spot the derivative of a product on the RHS. From there onwards, use integration.
3. (Original post by G A B R I E L)
I have:

that would go to:

Right? Not sure what to do next

Initial conditions are y = 0 at t = 0 and the constant is meant to equal

The answer is meant to be:

How did they get to this result?
Hi

You have an inhomogeneous linear first order DE
First solve the homogeneous part

from this

and itegrating

similarly as you wrote as a result with difference of that -bt will be bt at the exponent because -(-b) =b
But this is the homogeneous solution (Y).
Second: You should to find a particular solution (yp)
(Your final solution will be y=Y+yp)
One method tofind particular solution is substituting k(t) function
in the coefficient of C in the general solution
So
With differentiation substitute it the original equation and arrange to k'(t).
WIth integration you will get k(t) and from it yp
So y=Y+yp
The value of C in the final solution you will get from the initial conditions
and these give that

as you wrote it
4. (Original post by Farhan.Hanif93)
I'm not sure what you mean by 'that would go to', as I have no idea what you've done there.
he's just done a few steps in one go, it's right

where did i go wrong?
5. (Original post by Pheylan)
he's just done a few steps in one go, it's right

where did i go wrong?
the bit where you integrated e^-bt
6. (Original post by anshul95)
the bit where you integrated e^-bt
lol ****

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