y'(t)=y(y+1)^2,y(0)=1

Lösung

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

4

Schritte zur Lösung

Finde

y (y + 1)^{2}

gegeben

y (0) = 1

y (y + 1)^{2} = y (y + 1)^{2}

y : 1

(y + 1)^{2} : (1 + 1)^{2}

= 1 \cdot (1 + 1)^{2}

1 \cdot (1 + 1)^{2} = 4

= 4

\frac{d y}{d x} = \frac{35}{y ^{\frac{1}{6}} + 49 x ^{2} y ^{\frac{1}{6}}}

\frac{d y}{d x} y = - 6 x^{\frac{3}{2}}

\frac{d y}{d x} = (\frac{1 + y ^{2}}{1 + x ^{2}})

\frac{d y}{d x} = \frac{2 y ^{2}}{x}, y (9) = \frac{1}{49}

sin (2 x) d x + cos (3 y) d y = 0, y (\frac{π}{2}) = (\frac{π}{3})