y^{''}-2y^'=x+2e^x,y(0)=17,y^'(0)= 55/4

解答

y^{^{''}} - 2 y^{^{'}} = x + 2 e^{x}, y (0) = 17, y^{^{'}} (0) = \frac{55}{4}

y = 8 e^{2 x} + 11 + \frac{- x ^{2} - x}{4} - 2 e^{x}

求解步骤

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

求解线性方程:

y = 8 e^{2 x} + 11 + \frac{- x ^{2} - x}{4} - 2 e^{x}

y = 8 e^{2 x} + 11 + \frac{- x ^{2} - x}{4} - 2 e^{x}

\frac{\partial}{\partial y} (ln (1 + x^{2} + e^{y} + z))

in v erse l a pl a ce \frac{s ^{2} + 6 s + 2}{s ( s + 1 ) ^{2}}

y^{^{'}} = t y - 2 t + y - 2

l a pl a ce t r an s f or m 3 t^{3} - 2 t^{2} - 7 t + 2

\frac{\partial}{\partial x} (1^{x})