4y^{'''}+12y^{''}-16y^'-48y=0

解答

4 y^{^{'''}} + 12 y^{^{''}} - 16 y^{^{'}} - 48 y = 0

y = c_{1} e^{- 3 t} + c_{2} e^{- 2 t} + c_{3} e^{2 t}

求解步骤

4 y^{'''} (t) + 12 y^{''} (t) - 16 y^{'} (t) - 48 y = 0

求解线性方程:

y = c_{1} e^{- 3 t} + c_{2} e^{- 2 t} + c_{3} e^{2 t}

y = c_{1} e^{- 3 t} + c_{2} e^{- 2 t} + c_{3} e^{2 t}

t an g e n t f (x) = 8 x^{3} - 24 x^{2} - 193 x + 48

\frac{d}{d x} (sin (x) (x^{2} - 27 x + 6))

49 y^{^{''}} + 14 y^{^{'}} + y = 0

\frac{d}{d x} ((2 x^{2} + 1)^{\frac{3}{2}})

\int_{- 2}^{1} (8 x^{3} - 3 x^{2}) d x