2yy^'=18t^2,y(0)=5

Lösung

2 y y^{^{'}} = 18 t^{2}, y (0) = 5

y = 6 t^{3} + 25

Schritte zur Lösung

2 y y^{'} (t) = 18 t^{2}

Solve separable ODE:

y = 6 t^{3} + 25

y = 6 t^{3} + 25

6 y^{^{'}} + 5 y = 0

\frac{d y}{d x} sin (y) = (6 x^{4} - 6)

x y^{^{'}} - 2 y + 4 = 0

y^{^{'}} = \frac{( x ^{3} + x )}{4 y ^{3}}

y^{^{'}} = 0.002 \cdot (1000 - y), y (0) = 1