高数题怎么做
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f(x) 即如下分段函数:
f(x) = ax+b, 当 |x| < 1;
f(x) = 1/x , 当 |x| > 1;
f(x) = (a+b+1)/2, 当 x = 1;
f(x) = (b-a-1)/2, 当 x = -1.
间断点 x = 1, x = -1.
lim<x→-1->f(x) = lim<x→-1-> (1/x) = -1 = f(-1) = (b-a-1)/2;
lim<x→-1+>f(x) = lim<x→-1+> (ax+b) = b-a = f(-1) = (b-a-1)/2.
得 b-a = -1
lim<x→1->f(x) = lim<x→1-> (ax+b) = a+b = f(1) = (a+b+1)/2;
lim<x→1+>f(x) = lim<x→1+> (1/x) = 1 = f(1) = (a+b+1)/2.
得 a+b = 1,与 b-a = -1 联立解得 b = 0, a = 1
f(x) = ax+b, 当 |x| < 1;
f(x) = 1/x , 当 |x| > 1;
f(x) = (a+b+1)/2, 当 x = 1;
f(x) = (b-a-1)/2, 当 x = -1.
间断点 x = 1, x = -1.
lim<x→-1->f(x) = lim<x→-1-> (1/x) = -1 = f(-1) = (b-a-1)/2;
lim<x→-1+>f(x) = lim<x→-1+> (ax+b) = b-a = f(-1) = (b-a-1)/2.
得 b-a = -1
lim<x→1->f(x) = lim<x→1-> (ax+b) = a+b = f(1) = (a+b+1)/2;
lim<x→1+>f(x) = lim<x→1+> (1/x) = 1 = f(1) = (a+b+1)/2.
得 a+b = 1,与 b-a = -1 联立解得 b = 0, a = 1
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dy/dx +y = e^(-x)
let
y= (Ax+B)e^(-x)
y' =[-(Ax+B) +A].e^(-x) = [-Ax+(A-B)].e^(-x)
y'+y=e^(-x)
[-Ax+(A-B)].e^(-x) + (Ax+B)e^(-x) =e^(-x)
Ae^(-x) = e^(-x)
A=1
通解
y= (x+B)e^(-x)
let
y= (Ax+B)e^(-x)
y' =[-(Ax+B) +A].e^(-x) = [-Ax+(A-B)].e^(-x)
y'+y=e^(-x)
[-Ax+(A-B)].e^(-x) + (Ax+B)e^(-x) =e^(-x)
Ae^(-x) = e^(-x)
A=1
通解
y= (x+B)e^(-x)
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