y^{''}-y^'=e^{2x}*cos(e^x)

解答

y^{^{''}} - y^{^{'}} = e^{2 x} \cdot cos (e^{x})

y = c_{1} e^{x} + c_{2} - cos (e^{x})

求解步骤

y^{''} (x) - y^{'} (x) = e^{2 x} \cdot cos (e^{x})

求解线性方程:

y = c_{1} e^{x} + c_{2} - cos (e^{x})

y = c_{1} e^{x} + c_{2} - cos (e^{x})

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x^{2} + 3 + 2 cos^{2} (x)

y^{^{''}} - 3 y^{^{'}} + 10 y = cosh (2 x)

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

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

y^{^{''}} + 4 y = 0, y^{1} (0) = sin^{2} (x)