求(1+x)y''+y'=(1+x)^2满足初始条件y(0)=1,y'(0)=-1的特解
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求(1+x)y''''+y''=(1+x)^2满足初始条件y(0)=1,y''(0)=-1的特解求(1+x)y''''+y''=(1+x)^2满足初始条件y(0)=1,y''(0)=-1的特解求(1+x)y''''+y''
求(1+x)y''+y'=(1+x)^2满足初始条件y(0)=1,y'(0)=-1的特解
求(1+x)y''+y'=(1+x)^2满足初始条件y(0)=1,y'(0)=-1的特解
求(1+x)y''+y'=(1+x)^2满足初始条件y(0)=1,y'(0)=-1的特解
分离变量
dy/dx=[x(1+y^2)]/[(1+x^2)y]
把x,dx都挪到右边,y,dy挪到左边
ydy/(1+y^2)=xdx/(1+x^2)
两边积分
∫ydy/(1+y^2)=∫xdx/(1+x^2)
1/2∫d(1+y^2)/(1+y^2)=1/2∫d(1+x^2)/(1+x^2)
ln|1+y^2|=ln|1+x^2|+C'
e^ln(1+y^2)=e^[ln(1+x^2)+C']=e^C'[e^ln(1+x^2)] (能去绝对值因为1+x^2>0,1+y^2>0)
1+y^2=C(1+x^2)
代入x=0,y=1
1+1=C(1+0)
C=2
1+y^2=2(1+x^2)
y^2=2x^2+1
因为y(0)=1>0
所以开方
y=根号(2x^2+1) (舍去-根号(2x^2+1)
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