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

Solución

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

y = e^{t} (c_{1} cos (t) + c_{2} sin (t)) + 2 e^{t}

Pasos de solución

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

Resolver EDO lineales:

y = e^{t} (c_{1} cos (t) + c_{2} sin (t)) + 2 e^{t}

y = e^{t} (c_{1} cos (t) + c_{2} sin (t)) + 2 e^{t}

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

y^{^{''}} + y^{^{'}} + 2 y = x + 3, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + y^{^{'}} + 4.25 = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{- 5 x} (- 55 + 68 x - 30 x^{2})

2 x^{^{''}} + x = 3 t^{2} + 10 t