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

Lösung

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

y = tan (t + \frac{π}{4})

Schritte zur Lösung

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

Solve separable ODE:

y = tan (t + \frac{π}{4})

y = tan (t + \frac{π}{4})

im pl i c i t \frac{d y}{d x}, x = y e^{y}

n = 1 \sum \infty 3 (- \frac{2}{3})^{n}

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

\int \frac{x}{- x ^{2} + 5} d x

\int_{0}^{\frac{3}{2}} x cos (π x) d x