y-yy^'+2y=cos(8)

解答

y - y y^{^{'}} + 2 y = cos (8)

y = \frac{cos ( 8 ) W ( - \frac{1}{c o s ( 8 ) e ^{\frac{- 9 t - c _{1}}{c o s ( 8 )}}} )}{3} + \frac{cos ( 8 )}{3}

求解步骤

y - y y^{'} (t) + 2 y = cos (8)

求解可分离方程:

y = \frac{cos ( 8 ) W ( - \frac{1}{c o s ( 8 ) e ^{\frac{- 9 t - c _{1}}{c o s ( 8 )}}} )}{3} + \frac{cos ( 8 )}{3}

y = \frac{cos ( 8 ) W ( - \frac{1}{c o s ( 8 ) e ^{\frac{- 9 t - c _{1}}{c o s ( 8 )}}} )}{3} + \frac{cos ( 8 )}{3}

(t^{2} - 24 t + 95) y^{^{'}} = y

\frac{d}{d x} (x^{4} + y^{4}) = \frac{d}{d x} (16)

(x^{2} - 1) \frac{d y}{d x} = y^{2} - 1, y (2) = 2

\frac{d y}{d t} = k y + n

x \frac{d y}{d x} = e^{2 y + x}