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

Lösung

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

y = e^{- \frac{2 t ^{3}}{3} - t^{2} + c_{1}}

Schritte zur Lösung

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

Solve separable ODE:

y = e^{- \frac{2 t ^{3}}{3} - t^{2} + c_{1}}

y = e^{- \frac{2 t ^{3}}{3} - t^{2} + c_{1}}

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

\frac{1}{t} y^{^{'}} = 6 sin^{2} (y), y (1) = \frac{5 π}{4}