y^'= 1/2 y^3,y(1)=4

Lösung

y^{^{'}} = \frac{1}{2} y^{3}, y (1) = 4

y = - \frac{16}{16 t - 17}

Schritte zur Lösung

y^{'} (t) = \frac{1}{2} y^{3}

Solve separable ODE:

y = - \frac{16}{16 t - 17}

y = - \frac{16}{16 t - 17}

\frac{d y}{d x} = \frac{3 x ^{2}}{1 - 2 y}

x^{2} y^{^{'}} = y^{2}

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

x^{2} y^{^{'}} + \frac{1}{y ^{2}} = 0

y^{2} (x - 3) d x - x^{3} d y = 0