Question

高等数学，求解，谢谢

Answer 1

1. x^2dx - y(1+x^2)dy = 0, [x^2/(1+x^2)]dx = ydy,
[1-1/(1+x^2)]dx = ydy, x- arctanx + C/2 = y^2/2,
y^2 = 2x - 2arctanx + C.

1. y' + y = e^(-x), 一阶线性微分方程，
y = e^(-∫dx) [ ∫e^(-x)e^(∫dx) dx + C]
= e^(-x) (x+c)

2. 2xydy + (y^2-3x^2)dx = 0,
dy/dx = (3x^2-y^2)/(2xy) = (3/2)x/y - (1/2)y/x
是齐次方程， 令 y/x = u, 则 y = xu
u+du/dx = 3/(2u) - u/2, du/dx = (-3/2)(u^2-1)/u,
udu/(u^2-1) = (-3/2)dx
ln(u^2-1) = -3x + lnC
u^2-1 = Ce^(-3x)
y^2/x^2 - 1 = Ce^(-3x),
y(1) = √2 代入，得 1 = Ce^(-3√2), C = e^(3√2),
y^2/x^2 - 1 = e^(3√2-3x),