4y^{''}-y^'+2y=0,y(0)=2,y^'(0)=0

Lösung

4 y^{^{''}} - y^{^{'}} + 2 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y = e^{\frac{t}{8}} (2 cos (\frac{31 t}{8}) - \frac{2}{31} sin (\frac{31 t}{8}))

Schritte zur Lösung

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

Solve linear ODE:

y = e^{\frac{t}{8}} (2 cos (\frac{31 t}{8}) - \frac{2}{31} sin (\frac{31 t}{8}))

y = e^{\frac{t}{8}} (2 cos (\frac{31 t}{8}) - \frac{2}{31} sin (\frac{31 t}{8}))

x \to 3 lim (\frac{3 - x}{∣ 3 - x ∣})

x \to - \infty lim (0^{x})

x \to 0 + lim (ln (5) sin (x))

\int_{4}^{16} \frac{1}{6 - x} d x

im pl i c i t \frac{d y}{d x}, x y = 8