y^'+y=1,y(0)=0

Lösung

y^{^{'}} + y = 1, y (0) = 0

y = - e^{- t} + 1

Schritte zur Lösung

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

Solve separable ODE:

y = - e^{- t} + 1

y = - e^{- t} + 1

y^{^{'}} + 4 x (y^{2} + 2 y + 1) = 0

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

x y^{^{'}} = (1 - x) y

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

x^{^{'}} = \frac{1 + t}{t ^{2} x ^{2}}