y^{'''}-y^'=csc(t)

解答

y^{^{'''}} - y^{^{'}} = csc (t)

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} - ln (tan (\frac{t}{2})) + e^{- t} \cdot \int \frac{e ^{t} csc ( t )}{2} d t + e^{t} \cdot \int \frac{e ^{- t} csc ( t )}{2} d t

求解步骤

y^{'''} (t) - y^{'} (t) = csc (t)

求解线性方程:

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} - ln (tan (\frac{t}{2})) + e^{- t} \cdot \int \frac{e ^{t} csc ( t )}{2} d t + e^{t} \cdot \int \frac{e ^{- t} csc ( t )}{2} d t

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} - ln (tan (\frac{t}{2})) + e^{- t} \cdot \int \frac{e ^{t} csc ( t )}{2} d t + e^{t} \cdot \int \frac{e ^{- t} csc ( t )}{2} d t

y^{^{'''}} + 4 y^{^{'}} = sin (x) cos (x)

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

y^{^{''}} + 25 y = 9 sec (5 x)

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

y^{^{''}} + 9 y = g (t), y (0) = 2, y^{^{'}} (0) = 0