y^{''}+16y=5cos(4t),y(0)=1,y^'(0)=0

Lösung

y^{^{''}} + 16 y = 5 cos (4 t), y (0) = 1, y^{^{'}} (0) = 0

y = \frac{5 t sin ( 4 t )}{8} + cos (4 t)

Schritte zur Lösung

y^{''} (t) + 16 y = 5 cos (4 t)

Solve linear ODE:

y = \frac{5 t sin ( 4 t )}{8} + cos (4 t)

y = \frac{5 t sin ( 4 t )}{8} + cos (4 t)

y^{^{''''}} + 9 y^{^{''}} = 20 e^{- t}

y^{^{''''}} - y = 9 t + cos (t)

y^{^{''}} - y = 2.9 cos (t), y (0) = 0, y^{^{'}} (0) = 2.9

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} - 5 y = e^{x}

\frac{d ^{2} y}{d t ^{2}} - 4 \frac{d y}{d t} + 4 y = e^{- 4 t}