1. ## dy/dx = ky

another problem just want to check im doing it right?
....solve differential equation leaving unknown letter K in constant

$\frac{dy}{dx}=ky$

dy = ky dx

$y = \int ky + dx$

$=k\frac{y^2}{2} dx$

$\frac{1}{2}Ky^2 +C$

2. Originally Posted by decoy808
another problem just want to check im doing it right?
....solve differential equation leaving unknown letter K in constant

$\frac{dy}{dx}=ky$

dy = ky dx

$y = \int ky + dx$

$=k\frac{y^2}{2} dx$

$\frac{1}{2}Ky^2 +C$
Actually
$\displaystyle \frac{dy}{y} = k~dx$

Take it from here...

-Dan

y = Kx + C

?

4. No, you need to integrate the LHS as well.

$\ln |y| = kx + C$

5. thanks

6. As a side note, if we drop the "k", we are left with
dy/dx = y
In other words, there is a function whose derivative is the function itself.
Have you come across such a function in calculus?
(I hope so!)

Such a function is often used to model growth and decay...