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ODE question (integrating factor)

I have:

dydt−by=a\frac {dy}{dt}-by = a

that would go to:

ye∫−bdt=∫ae∫−bdtdt+Cye^{\int -b dt} = \int ae^{\int -b dt} dt + C

Right? Not sure what to do next

Initial conditions are y = 0 at t = 0 and the constant is meant to equal ab\frac {a}{b}

The answer is meant to be: y=ab(ebt−1) y = \frac {a}{b} (e^{bt} - 1)

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

dydt−by=a\frac {dy}{dt}-by = a

that would go to:

ye∫−bdt=∫ae∫−bdtdt+Cye^{\int -b dt} = \int ae^{\int -b dt} dt + C

Right? Not sure what to do next

Initial conditions are y = 0 at t = 0 and the constant is meant to equal ab\frac {a}{b}

The answer is meant to be: y=ab(ebt−1) y = \frac {a}{b} (e^{bt} - 1)

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 p(x)=e∫bdt=ebt+Cp(x)=e^{\int bdt} = e^{bt+C}.
Now multiply through your differential equation by p(x) and spot the derivative of a product on the RHS. From there onwards, use integration.
Reply 2
Avatar for ztibor
ztibor
10
Original post by G A B R I E L
I have:

dydt−by=a\frac {dy}{dt}-by = a

that would go to:

ye∫−bdt=∫ae∫−bdtdt+Cye^{\int -b dt} = \int ae^{\int -b dt} dt + C

Right? Not sure what to do next

Initial conditions are y = 0 at t = 0 and the constant is meant to equal ab\frac {a}{b}

The answer is meant to be: y=ab(ebt−1) y = \frac {a}{b} (e^{bt} - 1)

How did they get to this result?


Hi

You have an inhomogeneous linear first order DE
First solve the homogeneous part
dydt−by=0\frac{dy}{dt}-by=0
from this
dyy=bdt\frac{dy}{y}=bdt
and itegrating
lny=bt+c→Y=Cebtlny=bt +c \rightarrow Y=Ce^{bt}
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 yp=k(t)ebty_p=k(t)e^{bt}
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
y=ab(ebt−1)\displaystyle y=\frac{a}{b}\left( e^{bt}-1 \right)
as you wrote it
(edited 13 years ago)
Reply 3
Avatar for Pheylan
Pheylan
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

ye−bt=a∫e−bt dtye^{-bt}=a\int e^{-bt}\ dt

ye−bt=−abe−bt+cye^{-bt}=-abe^{-bt}+c

0=c−ab⇔c=ab0=c-ab\Leftrightarrow c=ab

ye−bt=ab−abe−btye^{-bt}=ab-abe^{-bt}

y=abebt−aby=abe^{bt}-ab

y=ab(ebt−1)y=ab(e^{bt}-1)

where did i go wrong?
Reply 4
Avatar for anshul95
anshul95
11
Original post by Pheylan
he's just done a few steps in one go, it's right

ye−bt=a∫e−bt dtye^{-bt}=a\int e^{-bt}\ dt

ye−bt=−abe−bt+cye^{-bt}=-abe^{-bt}+c

0=c−ab⇔c=ab0=c-ab\Leftrightarrow c=ab

ye−bt=ab−abe−btye^{-bt}=ab-abe^{-bt}

y=abebt−aby=abe^{bt}-ab

y=ab(ebt−1)y=ab(e^{bt}-1)

where did i go wrong?

the bit where you integrated e^-bt
Reply 5
Avatar for Pheylan
Pheylan
Original post by anshul95
the bit where you integrated e^-bt


lol ****

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