What is a particular solution to the differential equation #dy/dx=(y+5)(x+2)# with #y(0)=-1#?

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Answer 1 · 2016-07-09T19:45:04

#y = 4e^(x(x/2+2)) -5#

separate it first

you get

#1/(y+5) dy/dx = x+2#

So integrating
#int \ 1/(y+5) dy/dx \ dx = int x+2\ dx#

#= int \ 1/(y+5) \ dy = int x+2\ dx#

So

#= ln(y+5) = x^2/2+2x + C#

feeding in the IV #y(0) = -1#

#= ln(-1+5) = C = ln 4#

#implies ln(y+5) = x^2/2+2x + ln 4#

#implies ln((y+5)/4) = x^2/2+2x #

#y = 4e^(x(x/2+2)) -5#