What is the general solution of the differential equation ? # xdy=(y^2-3y+2) dx #

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Answer 1 · 2017-11-07T22:18:20

# y = (Ax-2)/(Ax-1) #

We have:

# xdy=(y^2-3y+2) dx #

This is a First Order separable DE, so we can "separate the variables", to get:

# int \ 1/(y^2-3y+2) \ dy = int \ 1/x \ dx#

The RHS can be integrated directly, and for the LHS we can use partial fraction decomposition:

# 1/(y^2-3y+2) -= 1/((y-1)(y-2)) #
# " " = A/(y-1) + B/(y-2) #
# " " = (A(y-2)+B(y-1))/((y-1)(y-2)) #

Leading to the identity:

# 1 -= A(y-2)+B(y-1) #

Where #A,B# are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put # y = 1 => 1 = -A => A=-1#
Put # y = 2 => 1=B=> B=1#

So we can now write:

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

Which we can now integrate to give:

# - ln |y-1| + ln|y-2| = ln|x| + lnA #
# :. ln ((y-2)/(y-1)) = lnAx #
# :. (y-2)/(y-1) = Ax #

Which we can re-arrange:

# y-2 = (y-1)Ax #

# :. y-2 = Axy-Ax #
# :. y-Axy = 2 -Ax #
# :. y(1-Ax) = 2 -Ax #

# :. y = (2 -Ax)/(1-Ax) #

# :. y = (Ax-2)/(Ax-1) #