y^{'''}-y^{''}+y^'+3y=0

解答

y^{^{'''}} - y^{^{''}} + y^{^{'}} + 3 y = 0

y = c_{1} e^{- t} + e^{t} (c_{2} cos (2 t) + c_{3} sin (2 t))

求解步骤

y^{'''} (t) - y^{''} (t) + y^{'} (t) + 3 y = 0

求解线性方程:

y = c_{1} e^{- t} + e^{t} (c_{2} cos (2 t) + c_{3} sin (2 t))

y = c_{1} e^{- t} + e^{t} (c_{2} cos (2 t) + c_{3} sin (2 t))

\frac{d y}{d t} = 3 y + y^{4}, y (0) = 1

\frac{d}{d x} (- 20 x^{3} + 36 x^{2} - 12 x - 4)

\frac{d ^{2}}{d x ^{2}} (9 z^{2} e^{z})

\int_{0}^{8} \frac{1}{64 + x ^{2}} d x

\frac{d}{d x} (3 x^{2} - 3 x + 4)