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

Lösung

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

y = \frac{t}{4} - \frac{1}{8} sin (2 t)

Schritte zur Lösung

y^{''} (t) + 4 y = t

Solve linear ODE:

y = \frac{t}{4} - \frac{1}{8} sin (2 t)

y = \frac{t}{4} - \frac{1}{8} sin (2 t)

\frac{d ^{2} y}{d x ^{2}} - 16 y = x e^{x}

y^{^{''}} + y = e^{- 2 π}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 6 y^{^{'}} + 8 y = x^{3} - x + 1

y^{^{''}} - y = e^{2 t} + 4 t \cdot e^{2 t} + 8 t^{2} \cdot e^{t}

9 y^{^{''}} + 4 y = 3 x + 12 e - x, y (0) = 1, y^{^{'}} (0) = 0