Solve #dy/dx = r-ky# ?

1 Answer
Steve M
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) #

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