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 writing lnA==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)