# Thread: separable differential equation dy/dx = y/x

1. ## separable differential equation dy/dx = y/x

Question is to solve $\frac{{dy}}{{dx}} = \frac{y}{x}$

I'm completely lost with these, this is as far as I get and then I'm not sure what to do. I get as far as this

$
\begin{array}{l}
\equiv \int {\frac{{dy}}{y} = \int {\frac{{dx}}{x}} } \\
\equiv \int {\frac{1}{y}dy = \int {\frac{1}{x}dx} } \\
\end{array}
$

i know they need to be integrated from here so

$
\ln y = \ln x + c
$

where c is my constant

the answer given in the text is
$
y = kx
$

2. Exponentiate both sides of the equation (raise the ln by e). Then you get y=x*e^(c). Because c is a constant and e is a number it can be absorbed and called k, thus you get y=kx

3. thanks Jake. so with respect to the $e^c$ , as it is just a number you are replacing is with another term to short note it? ie $k=e^c$

4. yep

5. Originally Posted by Craka
Question is to solve $\frac{{dy}}{{dx}} = \frac{y}{x}$

I'm completely lost with these, this is as far as I get and then I'm not sure what to do. I get as far as this

$
\begin{array}{l}
\equiv \int {\frac{{dy}}{y} = \int {\frac{{dx}}{x}} } \\
\equiv \int {\frac{1}{y}dy = \int {\frac{1}{x}dx} } \\
\end{array}
$

i know they need to be integrated from here so

$
\ln {\color{red}|} y {\color{red}|} = \ln {\color{red}|} x {\color{red}|} + c
$

where c is my constant

the answer given in the text is
$
y = kx
$
The red modulus brackets which you neglected and which I've added are essential. Failure to include them means that your equation is wrong.

6. Thanks I'd forgotten that.