题目
微分方程 (y - xy')/(x + yy') = 2 满足初值条件 y|_(x=1) = 1 的特解为____.A. arctan(x)/(y) + ln(x^2 + y^2)= (pi)/(4) + ln2B. arctan(x)/(y) - ln(x^2 + y^2)= (pi)/(4) + ln2C. arctan(y)/(x) + ln(x^2 + y^2)= (pi)/(4) + ln2D. arctan(y)/(x) - ln(x^2 + y^2)= (pi)/(4) + ln2
微分方程 $\frac{y - xy'}{x + yy'} = 2$ 满足初值条件 $y|_{x=1} = 1$ 的特解为____.
A. $\arctan\frac{x}{y} + \ln(x^2 + y^2)= \frac{\pi}{4} + \ln2$
B. $\arctan\frac{x}{y} - \ln(x^2 + y^2)= \frac{\pi}{4} + \ln2$
C. $\arctan\frac{y}{x} + \ln(x^2 + y^2)= \frac{\pi}{4} + \ln2$
D. $\arctan\frac{y}{x} - \ln(x^2 + y^2)= \frac{\pi}{4} + \ln2$
题目解答
答案
C. $\arctan\frac{y}{x} + \ln(x^2 + y^2)= \frac{\pi}{4} + \ln2$