x^3+y^2sqrt(x^4+4)y^'=0

Lösung

x^{3} + y^{2} x^{4} + 4 y^{^{'}} = 0

y = 3 - \frac{3}{2} x^{4} + 4 + c_{1}

Schritte zur Lösung

x^{3} + y^{2} x^{4} + 4 y^{'} (x) = 0

Solve separable ODE:

y = 3 - \frac{3}{2} x^{4} + 4 + c_{1}

y = 3 - \frac{3}{2} x^{4} + 4 + c_{1}

\int_{0}^{x} (e^{- t^{2}}) d t

\frac{d}{d x} (\frac{f ( x )}{g ( x ) - 2})

im pl i c i t \frac{d y}{d x}, 3 x y + 1 = 0

\int_{0}^{33} 3 (x - 1)^{- \frac{1}{5}} d x

x y \frac{d y}{d x} = 2 y^{2} - x^{2}