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Question

(dy/dx)=(xy-x+y-1)/(xy+x+2y+2) and (x^2y^2+x^2+y^2+1)dx=(xy+y)dy
fins the general solution of each

Answer 1

These are both separable differential equations. For the first,

dy/dx=[x(y-1)+1(y-1)]/[x(y+1)+2(y+1)]
dy/dx=[(x+1)(y-1)]/[(x+2)(y+1)]
(y+1)/(y-1) dy=(x+1)/(x+2)dx

Now integrate, rearrange and exponentiate:
y+2 ln |y-1|=x- ln |x+2|+C
ln |(y-1)^2 (x+2)|=x-y+C
(y-1)^2 (x+2)=C exp(x-y)

The second is similar:
(x^2+1)(y^2+1)dx=y(x+1)dy
(x^2+1)/(x+1)dx=y/(y^2+1)dy

Now integrate:
-2 (x+1)+1/2 (x+1)^2+2 ln |x+1|=1/2 ln (y^2+1)+C

Answer 2

Let y = u/v Then dy/dx = [V(dv/dx) - U(dv/dx) ] / V²

(1) dy/dx = [(xy+x+2y+2)(y -1) - (xy -x+y +1)(y-1)] /(xy + x + 2y + 2)²


= y² - 3
------------------------
(xy + x + 2y + 2)²



(2) dy/dx = [(xy+y)(2yx^2y+2x) - (x^2y²+x²+y²+1)(y)] / (xy + y)²


= 2(xy)^4y + y(x² + 2yx^2y + 2x - x^2y² - y² -1)
----------------------------------------------------------
y² (x² + 2x + 1 )

Answer 3

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