y^{''}+2y^'=3+4sin(2x)

解答

y^{^{''}} + 2 y^{^{'}} = 3 + 4 sin (2 x)

y = c_{1} + c_{2} e^{- 2 x} + \frac{3 x - sin ( 2 x ) - cos ( 2 x )}{2}

求解步骤

y^{''} (x) + 2 y^{'} (x) = 3 + 4 sin (2 x)

求解线性方程:

y = c_{1} + c_{2} e^{- 2 x} + \frac{3 x - sin ( 2 x ) - cos ( 2 x )}{2}

y = c_{1} + c_{2} e^{- 2 x} + \frac{3 x - sin ( 2 x ) - cos ( 2 x )}{2}

\frac{d}{d x} (\frac{x ^{2} - 8 x + 15}{x ^{2} + 3 x + 2})

e x p an d (x + a)^{4} (x - a)^{3}

\int a x^{2} + b x + c d x

\frac{\partial}{\partial y} (\frac{x - y}{x y + 3})

x \to 0 lim (x ln^{3} (1 + 3 x^{3}))