(d^2-1)y=sech(x)

解答

(d^{2} - 1) y = sech (x)

y = c_{1} e^{x} + c_{2} e^{- x} + e^{x} \cdot \int \frac{e ^{- x} sech ( x )}{2} d x + e^{- x} (- \frac{1}{2} e^{x} sech (x) + \frac{1}{2} sech (x) arcsech (sech (x)) + arcsin (\frac{2}{2} sech (x) + 1))

求解步骤

(d^{2} - 1) y = sech (x)

求解线性方程:

y = c_{1} e^{x} + c_{2} e^{- x} + e^{x} \cdot \int \frac{e ^{- x} sech ( x )}{2} d x + e^{- x} (- \frac{1}{2} e^{x} sech (x) + \frac{1}{2} sech (x) arcsech (sech (x)) + arcsin (\frac{2}{2} sech (x) + 1))

y = c_{1} e^{x} + c_{2} e^{- x} + e^{x} \cdot \int \frac{e ^{- x} sech ( x )}{2} d x + e^{- x} (- \frac{1}{2} e^{x} sech (x) + \frac{1}{2} sech (x) arcsech (sech (x)) + arcsin (\frac{2}{2} sech (x) + 1))

(d + 2) \cdot (d - 1)^{3} \cdot y = e^{x}

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 2 - 24 e^{x} + 40 e^{5} x

(d^{2} + 9) y = 2 x^{2}

(d^{5} - d) \cdot j = 8 sin (x) - 2 \cdot x

(d^{2} + 4) y = sin^{2} (2 x)