1. ## A differential equation

Hi, i would like some help with this differential equation.

$\begin{cases}
dy/dt +10y = 1 \\
y(1/10) = 2/10
\end{cases}$

This is a separable differential equation but how would you separate the variables in this case?

2. Originally Posted by Jones
Hi, i would like some help with this differential equation.

$\begin{cases}
dy/dt +10y = 1 \\
y(1/10) = 2/10
\end{cases}$

This is a separable differential equation but how would you separate the variables in this case?
$\frac{dy}{dt} = 1 - 10y \Rightarrow \frac{dt}{dy} = \frac{1}{1 - 10y}$.

3. Originally Posted by mr fantastic
$\frac{dy}{dt} = 1 - 10y \Rightarrow \frac{dt}{dy} = \frac{1}{1 - 10y}$.
Okay so the next step would be to integrate?

$y = \int \frac{1}{1-10y}$
$\rightarrow \frac {1}{10} ln(1-10y)$ ?

4. Originally Posted by Jones
Okay so the next step would be to integrate?

$y = \int \frac{1}{1-10y}$
$\rightarrow \frac {1}{10} ln(1-10y)$ ?
Sloppy notation, try something like:

$
\int\ dt=\int \frac{1}{1-10y}\;dy
$

so:

$
t=-\;\frac {1}{10} \ln(1-10y) +C
$

Now continue...

CB

5. Isn't there an easier way of doing it?

Maybe you could use the integrating factor $e^{\int 10t}$
and multiply both sides of the equation by that?

6. Originally Posted by Jones
Isn't there an easier way of doing it?

Maybe you could use the integrating factor $e^{\int 10t}$
and multiply both sides of the equation by that?
Yes an integrating factor will work, but that's not it. But with the correct integrating factor you will get the same answer, it just depends on what you consider an easier method.

CB

7. Originally Posted by CaptainBlack
Sloppy notation, try something like:

$
\int\ dt=\int \frac{1}{1-10y}\;dy
$

so:

$
t=-\;\frac {1}{10} \ln(1-10y) +C
$

Now continue...

CB
Right,
$
\frac {1}{10} \ln(1-10y) +C
$

Becomes, $e^{1-10y} * \frac{1}{10} \Longrightarrow e^{-10y} *\frac{e}{10}$
and $y(1/10) = 2/10$ so
$e^{-1} *\frac{e}{10} = \frac{2}{10}$

Hence C must be $= \frac{1}{10}$

8. Originally Posted by Jones
Right,
$
\frac {1}{10} \ln(1-10y) +C
$

Becomes, $e^{1-10y} * \frac{1}{10} \Longrightarrow e^{-10y} *\frac{e}{10}$
and $y(1/10) = 2/10$ so
$e^{-1} *\frac{e}{10} = \frac{2}{10}$

Hence C must be $= \frac{1}{10}$
That is confused, you need to work more clearly. I would do this:

$t=-\frac {1}{10} \ln(1-10y) +C$

so:

$-10t=\ln(1-10y)+C$

Now exponentiating both sides:

$c\;e^{-10t}=1-10y$

and so:

$y=\frac{1}{10}[1-c\;e^{-10t}]$

Now we apply the condition $y(1/10)=2/10$ :

$\frac{2}{10}=\frac{1}{10}[1-c\;e^{-1}]$

so $c=-e$

and so:

$y=\frac{1}{10}[1+e^{-10t+1}]$

CB

9. Hi,

Thank you. I understand all the algebra. The part that i don't really get is the
$\int \frac {dy}{dt}$ i think the notation is confusing. What do you end up with after integrating that?

10. Originally Posted by Jones
Hi,

Thank you. I understand all the algebra. The part that i don't really get is the
$\int \frac {dy}{dt}$ i think the notation is confusing. What do you end up with after integrating that?
At no point was this integrated, what was integrated was two different forms of:

$\int \frac {dt}{dy} \;dy$

(see Mr Fantastics post)

CB

11. Ok,

Just one last question.
If you have the equation $y' + 10y = 0$

The solution to this equation is $Ce^{-10t}$

What will the general solution be if you have longer expressions
for example $y' +3x^2y-x^2 = 0$

Is it simply e to the power of that polynomial ?

12. Originally Posted by Jones
Ok,

Just one last question.
If you have the equation $y' + 10y = 0$

The solution to this equation is $Ce^{-10t}$

What will the general solution be if you have longer expressions
for example $y' +3x^2y-x^2 = 0$

Is it simply e to the power of that polynomial ?
No, as you can verify for yourself by differentiating what you just suggested and substituting into the equation.

CB

13. Originally Posted by Jones
[snip]
What will the general solution be if you have longer expressions
for example $y' +3x^2y-x^2 = 0$

Is it simply e to the power of that polynomial ?
The DE can be re-written as $\frac{dy}{dx} = x^2 - 3x^2 y = x^2 (1 - 3y)$ which is readily seperable.

However, if it was $y' + 3y - x^2 = 0$, say .....

This is NOT seperable so now the easiest approach might be to re-write it as $y' + 3 y = x^2$ and use the integrating factor method.