2y^'+3y=0

Lösung

2 y^{^{'}} + 3 y = 0

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

Schritte zur Lösung

2 y^{'} (t) + 3 y = 0

Solve separable ODE:

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

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

d er i v a t i v e x \cdot 3 x^{2} + 5

a re a x^{2} - 2, x, - 2, 2

x \to - 4 lim (- x^{2} + 4 x - 9)

f (x) = ln (1 - x)

\int x^{8.3} d x