y^{''}+y=sec(x^3)

解答

y^{^{''}} + y = sec (x^{3})

y = c_{1} cos (x) + c_{2} sin (x) + cos (x) \cdot \int - sin (x) sec (x^{3}) d x + sin (x) \cdot \int cos (x) sec (x^{3}) d x

求解步骤

y^{''} (x) + y = sec (x^{3})

求解线性方程:

y = c_{1} cos (x) + c_{2} sin (x) + cos (x) \cdot \int - sin (x) sec (x^{3}) d x + sin (x) \cdot \int cos (x) sec (x^{3}) d x

y = c_{1} cos (x) + c_{2} sin (x) + cos (x) \cdot \int - sin (x) sec (x^{3}) d x + sin (x) \cdot \int cos (x) sec (x^{3}) d x

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

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

so l v e f or d, d (t) = 4000 - 3000 e^{- 0.06 t}

(d^{2} - d - 2) \cdot y = 2 \cdot lo g_{10} (x) + \frac{1}{x} + \frac{1}{( x ^{2} )}

(d - 5)^{2} y = e^{5 x}