x^'=4x+0x,x^'(0)=-4x+(-2)x

Lösung

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

x = e^{4 t + c_{1}}

Schritte zur Lösung

x^{'} (t) = 4 x + 0 x

Solve separable ODE:

x = e^{4 t + c_{1}}

x = e^{4 t + c_{1}}

y^{^{'}} = 7 x^{2} y^{- \frac{1}{2}}, y (1) = 1

y^{^{'}} = a (b - y) - cy

y^{^{'}} = a \cdot e^{- b \cdot y}

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

- 2 y^{^{'}} - 3 y = 0