What is the solution for (y^2-x^2)dx+(-2xy)dy=0 ?

The solution for (y^2-x^2)dx+(-2xy)dy=0 is y=sqrt(-x^2+c_{1)x},y=-sqrt(-x^2+c_{1)x}

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(y^2-x^2)dx+(-2xy)dy=0

(y^{2} - x^{2}) d x + (- 2 x y) d y = 0

y = - x^{2} + c_{1} x, y = - - x^{2} + c_{1} x

Solution steps

(y^{2} - x^{2}) d x + (- 2 x y) d y = 0

Solve Bernoulli ODE:

y = - x^{2} + c_{1} x, y = - - x^{2} + c_{1} x

y = - x^{2} + c_{1} x, y = - - x^{2} + c_{1} x

y^{^{''}} - 4 y^{^{'}} - 12 y = 8 e^{- 2 x}

\frac{d}{d x} (x e^{x} cos (x))

\int (5 x^{2} + 2 x^{3}) d x

x (\frac{d y}{d x}) = x tan (\frac{y}{x}) + y

\int (3 x + 5)^{6} d x

What is the solution for (y^2-x^2)dx+(-2xy)dy=0 ?
The solution for (y^2-x^2)dx+(-2xy)dy=0 is y=sqrt(-x^2+c_{1)x},y=-sqrt(-x^2+c_{1)x}