y^'=y^2+1,y(1)=0

Lösung

y^{^{'}} = y^{2} + 1, y (1) = 0

y = tan (t - 1)

Schritte zur Lösung

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

Solve separable ODE:

y = tan (t - 1)

y = tan (t - 1)

d y = θ^{3} d θ

\frac{d y}{d x} y = \frac{( 5 x ^{2} + 2 ) ( 3 x + 9 )}{8 x - 9}

3 e^{2 x} \frac{d y}{d x} = - 4 \cdot \frac{x}{y ^{2}}

\frac{d y}{d x} = \frac{( 1 - y )}{xv}

y^{^{'}} = 1 - \frac{y}{10}