yy^'=4x^3+2x

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

\frac{d V}{d t} = - V

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

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

2 z u d uu d u + u^{2} d z = 0

\frac{d x}{d t} = \frac{1}{x e ^{t + 9 x}}