Solve dy/dx = r-ky ?
1 Answer
Apr 20, 2017
y = r/k-Be^(-kx)
Explanation:
We have:
dy/dx = r-ky
Which is a first order separable Differential Equation. We can rearrange as follows
1/(r-ky)dy/dx = 1
So we can "separate the variables" to get:
int \ 1/(r-ky) \ dy = int \ dx
Integrating gives us:
-1/k ln(r-ky) = x + C
:. ln(r-ky) = -kx -kC
:. ln(r-ky) = -kx +ln A \ \ (by writinglnA==kC )
:. ln(r-ky) -lnA = -kx
:. ln((r-ky)/A) = -kx
:. (r-ky)/A = e^(-kx)
:. r-ky = Ae^(-kx)
:. ky = r-Ae^(-kx)
:. y = r/k-Be^(-kx)