How to solve differential equation dy/dx = k(c - y)?
Just wondering how you would solve this kind of a differential equation.
thanks in advanced <3
- ?Lv 78 years agoFavorite Answer
Assuming that k and c are just constants, the equation is separable, divide both sides by (c - y) and multiply the differential across:
1/(c - y) * dy = k * dx
Integrate both sides:
-ln(c - y) = kx + C
Make y the subject, divide both sides by -1 and raise both sides to the base of e:
c - y = e^(-kx) * e^-C
As e^-C represents a constant, we can just replace it with another constant placeholder:
c - y = C₁e^(-kx)
y = c - C₁e^(-kx)
- Demiurge42Lv 78 years ago
dy/dx = k(c - y)
dy/(c-y) = k dx for y≠c (because this would cause division by zero)
∫ dy/(c-y) = ∫k dx
-ln|c-y| = kx + c_1 where c_1 is a real constant
ln|c-y| = -kx + c_2 where c_2 = -c_1
|c-y| = e^(-kx + c_2) = c_3*e^(-kx) where c_3 = e^c_2 (c_3 is a positive number)
c-y = ±c_3*e^(-kx) = c_4 e^-(kx) where c_4 = ± c_3 (c_4 can't be zero)
y = -c_4 e^(-kx)+c
If we let c_4 = 0 we also get the valid solution y = c (you can check this by plugging it into the differential equation.)
y = m e^(-kx) + c where m is a real constant
- InfinityLv 48 years ago
dy/dx = k(c-y)
dy/(c-y) = kdx
Integrate both sides.
-ln(c-y) = kx + C
ln(1/(c-y)) = kx + C
You can then re-arrange as you wish.