y^{''}-y^'-20=0

解答

y^{^{''}} - y^{^{'}} - 20 = 0

y = c_{1} e^{t} + c_{2} - 20 t

求解步骤

y^{''} (t) - y^{'} (t) - 20 = 0

求解线性方程:

y = c_{1} e^{t} + c_{2} - 20 t

y = c_{1} e^{t} + c_{2} - 20 t

\frac{d ^{2} y}{d x ^{2}} - 9 y = e^{3 x} cos (2 x) + e^{2 x}

y^{^{''}} + 9 y = cos (t)

y^{^{''}} + 4 y = t^{3} + 3 e^{t}, y (0) = 0, y^{^{'}} (0) = 2

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

y^{^{''}} - 8 y^{^{'}} + 16 = 10 e^{4 x} sec (x)