1/2 q^{''}+10q^'+(1/(10^{-2)})q=200

解答

\frac{1}{2} q^{^{''}} + 10 q^{^{'}} + (\frac{1}{1 0 ^{- 2}}) q = 200

q (t) = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + 2

求解步骤

\frac{1}{2} q^{''} (t) + 10 q^{'} (t) + (\frac{1}{1 0 ^{- 2}}) q (t) = 200

求解线性方程:

q (t) = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + 2

q (t) = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + 2

y^{^{'''}} + 4 y^{^{''}} + 5 y^{^{'}} = 11 + 5 e^{2 x}

y^{^{''}} - 3 y^{^{'}} + 2 y = 3 e^{- x}

y^{^{''}} - 2 y^{^{'}} + y = e^{x} sech^{2} (x)

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 36 y = 4

y^{^{''}} + y = e^{3} + e^{2 x}