differential equations problem here! eliminate the arbitrary constant?

Question

pleae answr with solution.. tnx
1. xy^2-1=cy
2.cx^2+x+y^2=0

actually i got the answer here on my questionary book but i just dont know how it's solution. show me pls.

answer:

1.y^3dx+(xy^2+1)dy=0
2.y=xy'+(y')^2+1

Answer 1

1.
c=(xy^2-1)/y
difrentiate w.r.t x
(y(y^2 + 2xyp)-p(xy^2-1))/y^2=0
y^3 + 2xy^2p - pxy^2 + p=0
y^3 +xy^2p +p=0
where p=(dy/dx)
2.
similarly diffrentiate
(x+y^2)/x^2=-c
to get
2(x^2)yp - x^2 - 2xy^2=0
thanks

Answer 2

y = (c1*e^2t) + (c2*e^-3t) the coefficients of the powers of the exponents are the roots of the function equation. roots: from y = (c1*e^2t) + (c2*e^-3t) the roots may be desperate as r = 2 and r = -3 this is r - 2 = 0 and r + 3 = 0 so (r - 2)(r + 3) = 0 r² + r - 6 = 0 deduce the differential equation from the function equation y'' + y' - 6y = 0 NB: this physique of innovations is sensible in case you realize the belief maximum excellent to the final answer.

Answer 3

There aren't any differential equations in the question you have asked, so I'm guessing that these are your solutions to differential equations problems you've been given, and you want to know how to eliminate c from your answers.

That's possible only (I think) if you've been given some boundary conditions, e.g. when x = ..., then y = ---. If so, sub these in the equation and find the value of c. If not, I don't know what it's about.