What is the solution for (2y^2+4x^2)dx-xydy=0,y(1)=-2 ?

The solution for (2y^2+4x^2)dx-xydy=0,y(1)=-2 is y=xsqrt(-4+c_{1)x^2},y=-xsqrt(-4+c_{1)x^2}

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(2y^2+4x^2)dx-xydy=0,y(1)=-2

(2 y^{2} + 4 x^{2}) d x - x y d y = 0, y (1) = - 2

y = x - 4 + c_{1} x^{2}, y = - x - 4 + c_{1} x^{2}

Solution steps

(2 y^{2} + 4 x^{2}) d x - x y d y = 0

Solve Bernoulli ODE:

y = x - 4 + c_{1} x^{2}, y = - x - 4 + c_{1} x^{2}

y = x - 4 + c_{1} x^{2}, y = - x - 4 + c_{1} x^{2}

\frac{d}{d x} (x \cdot tan (x^{2}))

f (x) = x \cdot e^{- x}

\frac{d}{d x} (-)

\frac{d x}{d t} = t + x^{2} t

\int sin (z) d z

What is the solution for (2y^2+4x^2)dx-xydy=0,y(1)=-2 ?
The solution for (2y^2+4x^2)dx-xydy=0,y(1)=-2 is y=xsqrt(-4+c_{1)x^2},y=-xsqrt(-4+c_{1)x^2}