Question

二阶微分方程求通解

二阶微分方程求通解两道全做，谢谢

Answer 1

（1）xy' = x - y
xy' + y = x
两边同时除以x可得：y'+y/x=1,
这是个一阶线性方程,解决方法为常数变异法.
先求其对应的齐次方程y'+y/x=0的通解为y=C/x,
所以原方程的通解形式为y=C(x)/x,
将该式代入原方程中可得：C'(x)=x,所以C(x)=x^2/2+C,
所以原方程通解为y=x/2+C/x
（2）(1-x^2)y''-xy'=2
不显含y型,记y'=p,则y"=dp/dx=p',
原微分方程可化为
(1-x^2)p'-xp=2
p'-x/(1-x^2)p=2/(1-x^2)
公式法得
p=[e^(∫x/(1-x^2)dx][C1+∫2/(1-x^2)[e^(∫-x/(1-x^2)dx]dx]
=e^(-1/2)ln(1-x^2)[C1+∫{2/(1-x^2)e^[(1/2)ln(1-x^2)]}dx]
=(1-x^2)^(-1/2)[C1+∫{[2/(1-x^2)](1-x^2)^(1/2)}dx]
=(1-x^2)^(-1/2)[C1+∫{[2/(1-x^2)]^(1/2)dx]
=(1-x^2)^(-1/2)[C1+2arcsinx]
即dy/dx=(1-x^2)^(-1/2)[C1+2arcsinx]
∫dy=∫(1-x^2)^(-1/2)[C1+2arcsinx]dx
y=(1/2)∫[C1+2arcsinx]d(C1+2arcsinx)
得y=(1/4)(C1+2arcsinx)^2+C2