2y^'+10y=1

Lösung

2 y^{^{'}} + 10 y = 1

y = - \frac{e ^{- 5 t - c_{1}}}{10} + \frac{1}{10}

Schritte zur Lösung

2 y^{'} (t) + 10 y = 1

Solve separable ODE:

y = - \frac{e ^{- 5 t - c_{1}}}{10} + \frac{1}{10}

y = - \frac{e ^{- 5 t - c_{1}}}{10} + \frac{1}{10}

y^{^{'}} + 2 t (t + 1) y = 0

y d x + (3 + x) d y = 0

d y = 5 x^{4} (2 - y) d x

(x - 1) d y + (y - 2) d x = 0

y d x - x d y = 2 d y