What is the general solution of the differential equation ? xdy=(y^2-3y+2) dx xdy=(y2−3y+2)dx
1 Answer
y = (Ax-2)/(Ax-1) y=Ax−2Ax−1
Explanation:
We have:
xdy=(y^2-3y+2) dx xdy=(y2−3y+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
Put
y = 1 => 1 = -A => A=-1
Puty = 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)