cos(x)y^'= y/(ln(y)),y(0)=1

解答

cos (x) y^{^{'}} = \frac{y}{ln ( y )}, y (0) = 1

y = e^{2 l n (t a n (x) + s e c (x))}, y = e^{- 2 l n (t a n (x) + s e c (x))}

求解步骤

cos (x) y^{'} (x) = \frac{y}{ln ( y )}

求解可分离方程:

y = e^{2 l n (t a n (x) + s e c (x))}, y = e^{- 2 l n (t a n (x) + s e c (x))}

y = e^{2 l n (t a n (x) + s e c (x))}, y = e^{- 2 l n (t a n (x) + s e c (x))}

\frac{d y}{d x} = - \frac{4 \cdot + 3 y + 15}{2 \cdot + y + 7}

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

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

\frac{d y}{d x} - \frac{4}{x} y = 0

x y^{^{'}} - x^{2} y = 0