2f(t)+2f^'(t)=1

Lösung

2 f (t) + 2 f^{^{'}} (t) = 1

f (t) = - \frac{e ^{- t - c_{1}}}{2} + \frac{1}{2}

Schritte zur Lösung

2 f (t) + 2 f^{'} (t) = 1

Solve separable ODE:

f (t) = - \frac{e ^{- t - c_{1}}}{2} + \frac{1}{2}

f (t) = - \frac{e ^{- t - c_{1}}}{2} + \frac{1}{2}

x y - y \frac{d y}{d x} = 2 x, y (- 1) = 3

y^{^{'}} = \frac{8 y}{x}

\frac{d y}{d x} = - \frac{3}{8} x^{4} y^{4}

y y^{^{'}} + x e^{x^{2}} = 0

(1 - x^{2}) \frac{d y}{d x} + x y = x, y (0) = 2