-6y^{''}-5y^'-y=-(t^2+2t+2e^{2t})

解答

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

y = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{t}{3}} + t^{2} - 8 t + 28 + \frac{2}{35} e^{2 t}

求解步骤

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

求解线性方程:

y = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{t}{3}} + t^{2} - 8 t + 28 + \frac{2}{35} e^{2 t}

y = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{t}{3}} + t^{2} - 8 t + 28 + \frac{2}{35} e^{2 t}

0.75 y^{^{''}} + 50 y = 5 t^{2}

(D^{2} - 5 D + 6) y = x^{2} e^{3 x}

y^{^{''}} + 10 y^{^{'}} + 20 y = 50 sinh (5 x)

\frac{d ^{4} y}{d x ^{4}} + 3 \frac{d ^{2} y}{d x ^{2}} - 4 = 0

y^{^{''}} - y^{^{'}} = x + sin (x)