Question

大家帮我解析这个常微分方程，要详细的过程。尤其是右边怎么积分

d（x+y）/（x+y）=[p（t）+q（t）]dt
d(x-y)/(x-y)=[p(t)-q(t)]dt

Answer 1

定义e²=exp（2）即e^x=exp(x)

d(x+y)/(x+y)=[p（t）+q（t）]dt
d(x-y)/(x-y)=[p(t)-q(t)]dt
积分得
ln(x+y)=∫[p(t)+q(t)]dt
ln(x-y)=∫[p(t)-q(t)]dt

(x+y)=exp(∫[p(t)+q(t)]dt)
(x-y)=exp(∫[p(t)-q(t)]dt)

x=1/2{exp(∫[p(t)+q(t)]dt)+exp(∫[p(t)-q(t)]dt)}
y=1/2{exp(∫[p(t)+q(t)]dt)-exp(∫[p(t)-q(t)]dt)}