1/t y^'=6sin^2(y),y(1)=(5pi)/4

Lösung

\frac{1}{t} y^{^{'}} = 6 sin^{2} (y), y (1) = \frac{5 π}{4}

y = arccot (- 3 t^{2} + 4) + πn

Schritte zur Lösung

\frac{1}{t} y^{'} (t) = 6 sin^{2} (y)

Solve separable ODE:

y = arccot (- 3 t^{2} + 4) + πn

y = arccot (- 3 t^{2} + 4) + πn

y^{^{'}} = \frac{y \cdot cos ( x )}{1 - 2 y}

y^{^{'}} = \frac{x}{y}, y (1) = 0

y^{^{'}} x ln (x) - y = 0, y (2) = ln (4)

y^{^{'}} + y tan (x) = 0

e^{x} = y y^{^{'}}