y'+p(x)y=q(x)(q(x)不等于0）的通解为?

Question

y'+p(x)y=q(x)(q(x)不等于0）的通解为?

Answer 1

y' + p(x)•y = q(x)
积分因子 = [e^∫ p(x) dx],乘以方程两边
y'•[e^∫ p(x) dx] + [e^∫ p(x) dx]•p(x)•y = q(x)•[e^∫ p(x) dx]
y'•[e^∫ p(x) dx] + [e^∫ p(x) dx]'•y = q(x)•[e^∫ p(x) dx]
[y•e^∫ p(x) dx]' = q(x)•[e^∫ p(x) dx]
y•[e^∫ p(x) dx] = ∫ 【q(x)•[e^∫ p(x) dx]】 dx + C
y = [e^-∫ p(x) dx] • 〖∫ 【q(x)•[e^∫ p(x) dx]】 dx + C〗
括号太多了,留意点看吧.