What is the general solution of the differential equation ? xdy=(y^2-3y+2) dx xdy=(y23y+2)dx

1 Answer
Nov 7, 2017

y = (Ax-2)/(Ax-1) y=Ax2Ax1

Explanation:

We have:

xdy=(y^2-3y+2) dx xdy=(y23y+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)