4y+7y^'+2y^{''}=10*cos(4x)

Lösung

4 y + 7 y^{^{'}} + 2 y^{^{''}} = 10 \cdot cos (4 x)

y = c_{1} e^{\frac{( - 7 + 17 ) x}{4}} + c_{2} e^{\frac{( - 7 - 17 ) x}{4}} + \frac{5}{28} sin (4 x) - \frac{5}{28} cos (4 x)

Schritte zur Lösung

4 y + 7 y^{'} (x) + 2 y^{''} (x) = 10 \cdot cos (4 x)

Solve linear ODE:

y = c_{1} e^{\frac{( - 7 + 17 ) x}{4}} + c_{2} e^{\frac{( - 7 - 17 ) x}{4}} + \frac{5}{28} sin (4 x) - \frac{5}{28} cos (4 x)

y = c_{1} e^{\frac{( - 7 + 17 ) x}{4}} + c_{2} e^{\frac{( - 7 - 17 ) x}{4}} + \frac{5}{28} sin (4 x) - \frac{5}{28} cos (4 x)

y^{^{''}} - y^{^{'}} + y = sin (t) - 3 e^{2 t}

y^{^{''}} - 2 y^{^{'}} + y = 3 e^{t}

y^{^{''}} + 2 y^{^{'}} + y = x^{2} e^{x} cos (x)

y^{^{''}} - 5 y^{^{'}} + 4 y = e^{x}

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{3 x} - sin (x)