y^{''}+y^'+2y=x+3

Lösung

y^{^{''}} + y^{^{'}} + 2 y = x + 3

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} - 8 y = 2 cos (2 x) + (3 x + 2) e^{- 2 x}

2 y^{^{''}} - 4 y^{^{'}} + 4 y = e x sec (x)

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = 6 + e^{\frac{x}{2}}

y^{^{''}} - y = x e^{3 x}

y^{^{'''}} + 2 y^{^{''}} + 4 y^{^{'}} + 8 y = t^{2}