(x^2+9)u^'+2x\H(x)=0

解答

(x^{2} + 9) u^{^{'}} + 2 x H (x) = 0

u (x) = \int - \frac{2 H ( x ) x}{x ^{2} + 9} d x

求解步骤

(x^{2} + 9) u^{'} (x) + 2 x H (x) = 0

整理

u^{'} (x) : u^{'} (x) = - \frac{2 H ( x ) x}{x ^{2} + 9}

若

f^{^{'}} (x) = g (x)

则

f (x) = \int g (x) d x

u (x) = \int - \frac{2 H ( x ) x}{x ^{2} + 9} d x

u (x) = \int - \frac{2 H ( x ) x}{x ^{2} + 9} d x

y^{^{'}} tan (x) - y = 0

- 4 y^{^{'}} = 1 + x + y + x y

(8 + y) y^{^{'}} = sin (6 + x)

\frac{d x}{d t} = 14 - 0.56 x, t (0) = 130

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