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

解答

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

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

求解步骤

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

求解线性方程:

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

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

y = x^{2} + 3, (1, 4)

\frac{d}{d x} (\frac{x ^{p (x)}}{x ^{m} - a ^{m}})

\frac{\partial}{\partial x} (\frac{( x + y )}{y ^{2} - x})

\frac{d}{d x} (\frac{1 + csc ( x ^{7} )}{1 - cot ( x ^{7} )})

\frac{9}{x - 6}