(dy)/(dt)+3y=cos(2t),y(0)=-1

解答

\frac{d y}{d t} + 3 y = cos (2 t), y (0) = - 1

y = \frac{2 sin ( 2 t )}{13} + \frac{3 cos ( 2 t )}{13} - \frac{16}{13 e ^{3 t}}

求解步骤

\frac{d y}{d t} + 3 y = cos (2 t)

求解线性方程:

y = \frac{2 sin ( 2 t )}{13} + \frac{3 cos ( 2 t )}{13} - \frac{16}{13 e ^{3 t}}

y = \frac{2 sin ( 2 t )}{13} + \frac{3 cos ( 2 t )}{13} - \frac{16}{13 e ^{3 t}}

\int \frac{a + b sin ^{3} ( t )}{cos ^{2} ( t )} d t

\int - 32 x d x

\int_{0}^{2} \frac{1}{x ^{2} - 6 x + 5} d x

\frac{d}{d x} (arctan (x) - \frac{π}{4})

t an g e n t f (x) = 3 sin (x)