3y^{''}+48y=0,y(0)=4,y^'(0)=8

Lösung

3 y^{^{''}} + 48 y = 0, y (0) = 4, y^{^{'}} (0) = 8

y = 4 cos (4 t) + 2 sin (4 t)

Schritte zur Lösung

3 y^{''} (t) + 48 y = 0

Solve linear ODE:

y = 4 cos (4 t) + 2 sin (4 t)

y = 4 cos (4 t) + 2 sin (4 t)

y^{^{''''}} + 4 y^{^{'''}} + y^{^{''}} - 4 y^{^{'}} - 2 y = 0

\frac{d ^{2} f}{d x ^{2}} + \frac{df}{d x} = 0

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

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

l in e a r y^{^{'''}} + 19 y^{^{''}} - 14 y^{^{'}} - 266 y = 0