linear y^{''}-3y^'+2y=e^x

Lösung

linear

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

y = c_{1} e^{2 x} + c_{2} e^{x} - e^{x} x

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} - e^{x} x

y = c_{1} e^{2 x} + c_{2} e^{x} - e^{x} x

y^{^{''}} - y^{^{'}} + y = (2 e^{3 t} + 5 t)^{7}

\frac{d ^{2} y}{d x ^{2}} - 9 \frac{d y}{d x} + 5 y = x e^{x}

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 t^{2} + 4 t - 5

y^{^{''}} - y = 7.1 cos (t), y (0) = 0, y^{^{'}} (0) = 7.1