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

Question

Isabella Ackerman · Accepted Answer

#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#

Charles Anctil · Answer

undefined To find the particular solution to the differential equation dy/dx = (y + 5)(x + 2) with the initial condition y(0) = -1, you can separate variables and then integrate both sides with respect to their respective variables. After solving the integral equation, you can find the particular solution by substituting the initial condition.

After integrating and solving, the particular solution to the given differential equation with the initial condition y(0) = -1 is y = -5 + 5(x + 2)ln|x + 2|.