y^3-y^'+0.5y=0

Lösung

y^{3} - y^{^{'}} + 0.5 y = 0

y = \frac{0.5 e ^{c_{1} + t}}{1 - e ^{t + c_{1}}}, y = - \frac{0.5 e ^{c_{1} + t}}{1 - e ^{t + c_{1}}}

Schritte zur Lösung

y^{3} - y^{'} (t) + 0.5 y = 0

Solve separable ODE:

y = \frac{0.5 e ^{c_{1} + t}}{1 - e ^{t + c_{1}}}, y = - \frac{0.5 e ^{c_{1} + t}}{1 - e ^{t + c_{1}}}

y = \frac{0.5 e ^{c_{1} + t}}{1 - e ^{t + c_{1}}}, y = - \frac{0.5 e ^{c_{1} + t}}{1 - e ^{t + c_{1}}}

\frac{d y}{d x} = \frac{7 x + 5}{y - 3}, y (0) = 2

y^{^{'}} - (2) \cdot y = 2

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

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

4 y \frac{d y}{d x} = 6 x^{2}