y^'+k^2y=0

解答

y^{^{'}} + k^{2} y = 0

y = e^{- k^{2} t + c_{1}}

求解步骤

y^{'} (t) + k^{2} y = 0

求解可分离方程:

y = e^{- k^{2} t + c_{1}}

y = e^{- k^{2} t + c_{1}}

(1 + x^{3}) \frac{d y}{d x} = x^{2} y

\frac{d y}{d x} = \frac{x ( 3 + y ^{2} )}{y ( 2 + x ^{2} )}

\frac{d y}{d t} = 2 y t

8 N^{^{'}} + 10 t N^{5} = 0

4 x y \frac{d y}{d x} = 1, y > 0