y^'=3cos(t)-2y,y(0)=0

解答

y^{^{'}} = 3 cos (t) - 2 y, y (0) = 0

y = \frac{3 ( sin ( t ) + 2 cos ( t ) )}{5} - \frac{6}{5 e ^{2 t}}

求解步骤

y^{'} (t) = 3 cos (t) - 2 y

求解线性方程:

y = \frac{3 ( sin ( t ) + 2 cos ( t ) )}{5} - \frac{6}{5 e ^{2 t}}

y = \frac{3 ( sin ( t ) + 2 cos ( t ) )}{5} - \frac{6}{5 e ^{2 t}}

\int 5 x e^{3 x} d x

t an g e n t f (x) = 5 x^{3} - 5 x, (1, 0)

\frac{d}{d x} (ln (cos (\frac{x - 1}{x})))

\frac{d}{d x} (\frac{cos ( 2 x )}{sin ( x )})

\int_{1}^{2} \frac{x}{x ^{2} + 4 x + 20} d x