solve: x^2 dx+y•(x-1) dy=0

Question

x^2 dx+y•(x-1) dy=0

Comments

x^2 dx+y•(x-1) dy=0

Which can be written dy/dx=1/y • x^2/1-x
and y•dy=x^2/1-x •dx
For the integral on the right we set 1-x _= t and dx = -dt

{x^2/1-x dx → -{1/t(1-t)^2 dt=- lnt+2t-t^2/2
that is : 1/2 y^2=2(1-x) -1/2(1-x)^2 - ln(1-x)+c power ~

y^2=(x+3)•(1-x)- ln(1-x)^2 +c power ~

(x+3)•(1-x) _= 3-2x-x^2

y=√Ln{C base 0/(1-x)^2} -x(x+2)

Answer 1

Solve: x^2∙dx+y(x+1)∙dy=0

Rewrite the ODE as, P∙dx+Q∙dy=0

Where P=x^2,   P_y=0.
And Q=y(x+1),     Q_x=y.

Since P_y≠Q_x, then the ODE is not exact.

IF – integration factor
h(x)=(P_y-Q_x)/Q=(0-y)/y(x+1) =(-1)/(x+1)

Since h(x) is a function of x only, then we have found an integrating factor, which is μ(x), where

μ(x)= e^(∫h(x)  dx)=e^(-∫dx/(x+1))=e^(-ln(x+1) )=1/(x+1)
μ(x)=1/(x+1)

The ODE can now be made exact as,

P'∙dx+Q'∙dy=dU(x,y)=∂U/∂x∙dx+∂U/∂y∙dy

And dU=0, which implies U(x,y)=const.

∂U/∂x=P'=μP=x^2/(x+1)
Using IBP, U(x,y)=1/2∙x^2-x+ln(x+1)+g(y)

∂U/∂y=Q^'=μQ=y
U(x,y)=1/2∙y^2+k(x)

Comparison of the two forms of U(x,y) gives us,
g(y)=1/2∙y^2,       k(x)=1/2∙x^2-x+ln(x+1)

Hence U(x,y)=1/2∙y^2+1/2∙x^2-x+ln(x+1)=const=K/2

y^2=K-x^2+2x-2 ln(x+1)