y^'=1+((y)^2)/4 ,y(0)=0.5

Lösung

y^{^{'}} = 1 + \frac{( y ) ^{2}}{4}, y (0) = 0.5

y = 2 tan (\frac{t + 2 \cdot 0.24497 \dots}{2})

Schritte zur Lösung

y^{'} (t) = 1 + \frac{( y ) ^{2}}{4}

Solve separable ODE:

y = 2 tan (\frac{t + 2 \cdot 0.24497 \dots}{2})

y = 2 tan (\frac{t + 2 \cdot 0.24497 \dots}{2})

yy (x + 1) + y^{^{'}} = 0

y^{^{'}} = - 8 y + 16

\frac{d y}{d x} = 0.5 (y e^{x})^{\frac{1}{3}}

y^{^{'}} = 6 y \cdot ln (\frac{2}{y})

u^{^{'}} = 2 x H (x)