Hi, i would like some help with this differential equation.
$\displaystyle \begin{cases}
dy/dt +10y = 1 \\
y(1/10) = 2/10
\end{cases}$
This is a separable differential equation but how would you separate the variables in this case?
Hi, i would like some help with this differential equation.
$\displaystyle \begin{cases}
dy/dt +10y = 1 \\
y(1/10) = 2/10
\end{cases}$
This is a separable differential equation but how would you separate the variables in this case?
Right,
$\displaystyle
\frac {1}{10} \ln(1-10y) +C
$
Becomes, $\displaystyle e^{1-10y} * \frac{1}{10} \Longrightarrow e^{-10y} *\frac{e}{10}$
and $\displaystyle y(1/10) = 2/10$ so
$\displaystyle e^{-1} *\frac{e}{10} = \frac{2}{10}$
Hence C must be $\displaystyle = \frac{1}{10}$
That is confused, you need to work more clearly. I would do this:
$\displaystyle t=-\frac {1}{10} \ln(1-10y) +C$
so:
$\displaystyle -10t=\ln(1-10y)+C$
Now exponentiating both sides:
$\displaystyle c\;e^{-10t}=1-10y$
and so:
$\displaystyle y=\frac{1}{10}[1-c\;e^{-10t}]$
Now we apply the condition $\displaystyle y(1/10)=2/10$ :
$\displaystyle \frac{2}{10}=\frac{1}{10}[1-c\;e^{-1}]$
so $\displaystyle c=-e$
and so:
$\displaystyle y=\frac{1}{10}[1+e^{-10t+1}]$
CB
Ok,
Just one last question.
If you have the equation $\displaystyle y' + 10y = 0$
The solution to this equation is $\displaystyle Ce^{-10t}$
What will the general solution be if you have longer expressions
for example $\displaystyle y' +3x^2y-x^2 = 0$
Is it simply e to the power of that polynomial ?
The DE can be re-written as $\displaystyle \frac{dy}{dx} = x^2 - 3x^2 y = x^2 (1 - 3y)$ which is readily seperable.
However, if it was $\displaystyle y' + 3y - x^2 = 0$, say .....
This is NOT seperable so now the easiest approach might be to re-write it as $\displaystyle y' + 3 y = x^2$ and use the integrating factor method.