5.求解初值问题 (x-lny)dy+ylnydx=0, y|x=1=e.

Question

如题所述

Answer 1

(x-lny)dy + ylnydx = 0, 两边同除以 ylnydy， 得
(x-lny)/(ylny) + dx/dy = 0 即
dx/dy + x/(ydy) = 1/y 为 x 对 y 的一阶线性微分方程， 通解是
x = e^[∫-dy/(ylny)] {∫(1/y)e^[∫dy/(ylny)]dy + C}
= e^[∫-dy/(ylny)] {∫(1/y)e^[∫dy/(ylny)]dy + C}
= (1/lny) {∫(1/y)lnydy + C} = (1/lny)[(1/2)(lny)^2 + C]
= (1/2)lny + C/lny
y(1) = e 代入， 得 1 = 1/2 + C，得 C = 1/2
特解 2x = lny + 1/lny

Answer 2

解:微分方程为(x-lny)dy+ylnydx=0，化为
x-lny+ylnydx/dy=0，x/y+lny×dx/dy=lny/(y)，d(x/y)/dy=lny/(y)，x/y=lny/(y)+c (c为任意常数)，微分方程的通解为x=lny+cy
∵y(1)=e ∴有1=1+ce，得:c=0 ∴微分方程的通解为y=eˣ本回答被网友采纳