y^{''}+2y^'+4y=6cos(3x)

Lösung

y^{^{''}} + 2 y^{^{'}} + 4 y = 6 cos (3 x)

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{36}{61} sin (3 x) - \frac{30}{61} cos (3 x)

Schritte zur Lösung

y^{''} (x) + 2 y^{'} (x) + 4 y = 6 cos (3 x)

Solve linear ODE:

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{36}{61} sin (3 x) - \frac{30}{61} cos (3 x)

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{36}{61} sin (3 x) - \frac{30}{61} cos (3 x)

x^{^{''}} + 4 x^{^{'}} + 3 x = 1, x^{^{'}} (0) = 0

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

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

y^{^{'''}} - y^{^{''}} - 4 y^{^{'}} + 4 y = 10 e^{3 t}

y^{^{''}} - 4 y^{^{'}} + 4 y = 4 x ln (x)