t*y^'+4y=8t

Lösung

t \cdot y^{^{'}} + 4 y = 8 t

y = \frac{8 t}{5} + \frac{c _{1}}{t ^{4}}

Schritte zur Lösung

t \cdot y^{'} (t) + 4 y = 8 t

Solve linear ODE:

y = \frac{8 t}{5} + \frac{c _{1}}{t ^{4}}

y = \frac{8 t}{5} + \frac{c _{1}}{t ^{4}}

y^{^{''}} + 3 y^{^{'}} - 4 y = e^{- 4 x} + x e^{- x}

\frac{\partial}{\partial x} (z^{x y^{2}})

x \to 0 + lim (\frac{x - ∣ x ∣}{∣ x ∣})

\frac{d y}{d t} = 2 t e^{- y}, y (0) = - 2

y^{^{''}} = 4 y, y (0) = 0, y^{^{'}} (0) = 1