1. First Order DE

Hi,

I am looking to solve the following First Order differential equation, and am not sure what method to use i.e. seperating variables or Integrating factor.

The equation is dw/dt = k(a-w)

The result that the book gives is:

w = a(1-Be^-kt)

Can some one please show me how this can be solved in full and what method to use.

Many thanks,

2. Originally Posted by aharris1
Hi,

I am looking to solve the following First Order differential equation, and am not sure what method to use i.e. seperating variables or Integrating factor.

The equation is dw/dt = k(a-w)

The result that the book gives is:

w = a(1-Be^-kt)

Can some one please show me how this can be solved in full and what method to use.

Many thanks,

$\displaystyle \frac{dw}{dt} = k(a-w)$ is seperable:

$\displaystyle \int \frac{1}{a - w} \, dw = \int k \, dt$.

Now integrate each side (don't forget the arbitrary constant) and then make w the subject.

3. Hi Mr Fantastic

Integrating should give:

ln(a-w)= kt +C

taking logs of both sides:

a-w = e^kt +e^c

Where do I go from here? How do I deal with the RHS of the equation?

Many thanks

4. Originally Posted by aharris1
Hi Mr Fantastic

Integrating should give:

ln(a-w)= kt +C

taking logs of both sides:

a-w = e^kt +e^c

Where do I go from here? How do I deal with the RHS of the equation?

Many thanks

$\displaystyle ln|a-w|=kt+c \iff a-w = e^{kt+c}$

$\displaystyle a-w=e^{kt+c} \underbrace{=e^{kt} \cdot e^{c}}_{exp Properties}$

$\displaystyle a-w=e^{c}e^{kt}$

isolating w

$\displaystyle w=a-e^{c}e^{kt}$ now we factor out an "a"

$\displaystyle w=a(1-\frac{e^c}{a}e^{kt})$

now relabeling $\displaystyle \frac{e^c}{a}$ as B

we get... $\displaystyle w=(1-Be^{kt})$

Yeah!!!

5. First Order DE

Hi The Empty Set

I have never seen factoring like that! How does that work?

I was going to multiply through by -1 to isolate w i.e.

w-a = -Be^-kt setting e^c = B,a constant

so:

w = a-Be^-kt

Regards,

6. Originally Posted by aharris1
w = a-Be^-kt
Now divide both sides by a and see what you get.
(This is the same as factorizing)

7. First Order DE

Hi Wingless

w/a = (a- Be^-kt)/a

w/a = 1-Be^-kt

w = a(1-Be^-kt) .... Right?

8. Originally Posted by aharris1
Hi Wingless

w/a = (a- Be^-kt)/a

w/a = 1-Be^-kt

w = a(1-Be^-kt) .... Right?
Well, the thing is,
$\displaystyle \frac{a- B e^{-kt}}{a}$

$\displaystyle \frac{a}{a} - \frac{B e^{-kt}}{a}$

$\displaystyle 1 - \frac{B}{a}. e^{-kt}$

Now, B/a will not be B anymore. We can call it something else, K for example.

$\displaystyle \frac{\omega}{a} = 1 - K. e^{-kt}$

$\displaystyle \omega = a(1 - K. e^{-kt})$

I'm not sure where your fault is. You probably forgot to divide both a and Be^(-kt) by a, or you probably forgot to give B/a a new letter.

9. Originally Posted by aharris1
Hi The Empty Set

I have never seen factoring like that! How does that work?

I was going to multiply through by -1 to isolate w i.e.

w-a = -Be^-kt setting e^c = B,a constant

so:

w = a-Be^-kt

Regards,

$\displaystyle w=a-e^{c}e^{-kt} \iff w=a-\frac{a}{a}e^c e^{-kt}$
$\displaystyle w=a-\frac{a}{a}e^c e^{-kt}=a(1-\frac{e^c}{a}e^{-kt})=a(1-Be^{-kt})$