y^'=((3x^2))/(2y)

Lösung

y^{^{'}} = \frac{( 3 x ^{2} )}{2 y}

y = x^{3} + c_{1}, y = - x^{3} + c_{1}

Schritte zur Lösung

y^{'} (x) = \frac{( 3 x ^{2} )}{2 y}

Solve separable ODE:

y = x^{3} + c_{1}, y = - x^{3} + c_{1}

y = x^{3} + c_{1}, y = - x^{3} + c_{1}

\frac{d}{d x} (4 x^{3} + 2 y^{5}) = 6

t^{2} y + 11 t y^{^{'}} - 11 y = 0, y^{1} (0) = t

y^{^{'}} = \frac{x y}{1 + x ^{2}}

(x^{2} + 9) u^{^{'}} + 2 x H (x) = 0

y^{^{'}} tan (x) - y = 0