y^{''}+y=sqrt(2sin(\sqrt{2t))}

Lösung

y^{^{''}} + y = 2 sin (2 t)

y = c_{1} cos (t) + c_{2} sin (t) + cos (t) \cdot \int - 2 sin (t) sin (2 t) d t + sin (t) \cdot \int 2 cos (t) sin (2 t) d t

Schritte zur Lösung

y^{''} (t) + y = 2 sin (2 t)

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) + cos (t) \cdot \int - 2 sin (t) sin (2 t) d t + sin (t) \cdot \int 2 cos (t) sin (2 t) d t

y = c_{1} cos (t) + c_{2} sin (t) + cos (t) \cdot \int - 2 sin (t) sin (2 t) d t + sin (t) \cdot \int 2 cos (t) sin (2 t) d t

y^{^{'''}} - 2 y^{^{''}} - 4 y^{^{'}} + 8 = 0

y^{^{''}} - 8 y^{^{'}} - 9 y = 2 e^{- x} - 9 x

y^{^{''}} - 5 y^{^{'}} + 4 y = x, y^{1} (0) = e^{x}

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e^{- x}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 5 y = e^{t}