y^{''}+6y^'-40y=1-3x+x^2

Lösung

y^{^{''}} + 6 y^{^{'}} - 40 y = 1 - 3 x + x^{2}

y = c_{1} e^{4 x} + c_{2} e^{- 10 x} - \frac{x ^{2}}{40} + \frac{27 x}{400} - \frac{129}{8000}

Schritte zur Lösung

y^{''} (x) + 6 y^{'} (x) - 40 y = 1 - 3 x + x^{2}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- 10 x} - \frac{x ^{2}}{40} + \frac{27 x}{400} - \frac{129}{8000}

y = c_{1} e^{4 x} + c_{2} e^{- 10 x} - \frac{x ^{2}}{40} + \frac{27 x}{400} - \frac{129}{8000}

y^{^{''}} + 10 y^{^{'}} + 25 y = e^{2 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 6 y^{^{'}} + 45 y = 6, y (0) = 5, y^{^{'}} (0) = 9

y^{^{''}} + 5 y^{^{'}} + 4 y = 3 e^{- 5 x}, y (0) = 5, y^{^{'}} (0) = 3

y^{^{''}} - 4 y^{^{'}} + 3 y = 5 e^{3 t}

y^{^{'''}} - y^{^{''}} - 9 y^{^{'}} + 9 y = 4 - e^{x} + e^{3 x}