作业帮高数求通解y’=(y^2-x)/2y(x-1）
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高数求通解
y’=(y^2-x)/2y(x-1）
y’=(y^2-x)/2y(x-1）
由原方程进行变换：
2y*y'=(y^-x)/(x-1)
2y*dy/dx=(y^-x)/(x-1)
d(y^)/dx=(y^-x)/(x-1)
令t=y^,于是原方程转化为关于t和x的微分方程：
dt/dx=(t-x)/(x-1)
dt/dx +[-1/(x-1)]*t=-x/(x-1)
此为一阶线性非齐次方程
其中,p(x)=-1/(x-1),q(x)=-x/(x-1)
套用公式：
∫p(x)dx=∫-dx/(x-1)=-ln(x-1)
于是,e^[-∫p(x)dx]=e^[ln(x-1)]=x-1
e^[∫p(x)dx]=e^[-(x-1)]=1/(x-1)
∫q(x)*e^[∫p(x)dx]
=∫[-x/(x-1)]*[1/(x-1)]dx=∫[-x/(x-1)^]dx
=-∫[(x-1)+1]dx/(x-1)^
=-∫dx/(x-1) - ∫dx/(x-1)^
=-ln(x-1) + 1/(x-1)
=1/(x-1) - ln(x-1)
于是,可求出
t=(x-1)*[1/(x-1) - ln(x-1) +C]
=1+C(x-1)-(x-1)ln(x-1)
于是：
y^=1+c(x-1)-(x-1)ln(x-1)
2y*y'=(y^-x)/(x-1)
2y*dy/dx=(y^-x)/(x-1)
d(y^)/dx=(y^-x)/(x-1)
令t=y^,于是原方程转化为关于t和x的微分方程：
dt/dx=(t-x)/(x-1)
dt/dx +[-1/(x-1)]*t=-x/(x-1)
此为一阶线性非齐次方程
其中,p(x)=-1/(x-1),q(x)=-x/(x-1)
套用公式：
∫p(x)dx=∫-dx/(x-1)=-ln(x-1)
于是,e^[-∫p(x)dx]=e^[ln(x-1)]=x-1
e^[∫p(x)dx]=e^[-(x-1)]=1/(x-1)
∫q(x)*e^[∫p(x)dx]
=∫[-x/(x-1)]*[1/(x-1)]dx=∫[-x/(x-1)^]dx
=-∫[(x-1)+1]dx/(x-1)^
=-∫dx/(x-1) - ∫dx/(x-1)^
=-ln(x-1) + 1/(x-1)
=1/(x-1) - ln(x-1)
于是,可求出
t=(x-1)*[1/(x-1) - ln(x-1) +C]
=1+C(x-1)-(x-1)ln(x-1)
于是：
y^=1+c(x-1)-(x-1)ln(x-1)