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

解答

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

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{ln ( sin ( 2 x ) ) - cos ( 2 x ) ( ln ( tan ( x ) ) + cos ( 2 x ) ) - sin ^{2} ( 2 x )}{8}

求解步骤

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

求解线性方程:

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{ln ( sin ( 2 x ) ) - cos ( 2 x ) ( ln ( tan ( x ) ) + cos ( 2 x ) ) - sin ^{2} ( 2 x )}{8}

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{ln ( sin ( 2 x ) ) - cos ( 2 x ) ( ln ( tan ( x ) ) + cos ( 2 x ) ) - sin ^{2} ( 2 x )}{8}

t an g e n t f (x) = x^{2}, (0, - 4)

d er i v a t i v e 3 x^{3} + \frac{4}{x}

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

y^{^{''}} + 9 y = - sin (3 x)

\int_{0}^{1} \frac{1}{x ^{2} - 2 x + 2} d x