y'= y^2+x^2 / 2xy; CDV: y=ux; dy=xdu + udx;

dy/dx = y^2+x^2 / 2xy; reemplazo:

(xdu + udx) /dx = [(ux)^2 + x^2] / 2ux^2; factor común a la derecha:

(xdu + udx) /dx = x^2*(u^2 + 1) / 2ux^2; simplifico a la derecha:

x(du/dx) + u = (u^2 + 1) / 2u;

x(du/dx)= [(u^2 + 1) / 2u] - u

x(du/dx)= [(u^2 - 2u^2 + 1) / 2u];

x(du/dx)= [(1-u^2) / 2u];

2u*du / (1-u^2) = dx/x; a la izquierda por fracciones parciales:

2u / [(1+u)(1-u)];

2u / [(1+u)(1-u)] = A/(1+u) + B/(1-u); denominador común a la derecha:

2u / [(1+u)(1-u)] = [A(1-u) + B(1-u)] / [(1+u)(1-u)]; simplifico:

2u = A(1-u) + B(1-u); genero matriz:

1A + 1B = 0; para u=0;

0A + 2B = 2; Para u=1; B=1; A= (-1);

du/(1-u) - du/(1+u) = dx/x; integro:

- ln|1-u| - ln|1+u| = ln|x| + C; o, si hago: ln|A|=C:

- ln|1-u| - ln|1+u| = ln|x| + ln|A|;

-ln |1-u^2| = ln |Ax|;

ln |1/(1-u^2)| = ln |Ax|

1/(1-u^2) = Ax;

1= Ax(1-u^2); devuelvo variable:

1= Ax [1 - (y/x)^2];

1= A (x^2 - y^2)/x;

x/A= x^2 - y^2;

y^2 = x^2 - (x/A);

y = +-√[x^2 - (x/A)]; Podemos hacer: A= 1/K:

y = +-√(x^2 - Kx).

Hola

v = y/x

y = v x

dy = v dx + x dv

y^2 + x^2 = v^2 x^2 + x^2 = x^2(1 + v^2 )

2 x y = 2 x v x = x^2 (2 v)

(y^2 + x^2)/(2 x y) = (1 + v^2)/(2 v)

queda

v' x + v = (1 + v^2)/(2 v)

v' x = (1 + v^2)/(2 v) - v

v' x = (1 + v^2 - 2 v^2)/(2 v)

v' x = (1 - v^2)/(2 v)

v dv/(1 - v^2) = dx/ x

- 2 v dv/(1 - v^2) = -2 dx/x

d(1 - v^2)/(1 - v^2) + 2 dx/x = 0

d ( ln(1 - v^2) + 2 ln(x) ) = 0

integramos

ln ( (1 - v^2) * x^2 ) = k1

x^2 - v^2 x^2 = k2 (= e^k1)

x^2 - y^2 = k2

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Saludos

PEZ, LA RESPUESTA SIEMPRE ES PEZ!