Question

高数求助啊。。

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Answer 1

(3)
let
u=y/x^4
du/dx = (1/x^4). dy/dx - 4y/x^5
dy/dx = x^4.[ du/dx + (4/x)u]
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dy/dx - (4/x)y = x√y
x^4.[ du/dx + (4/x)u] - (4/x^3)u = x^3.√u
x^4. du/dx =x^3.√u
x.du/dx =√u
∫ du/√u = ∫ dx/x
2√u = ln|x| +C'
2√y/x^2 = ln|x| +C'
y = (1/4) x^4 . ( ln|Cx| )^2
(6)
f(x) - ∫(0->x) (x-u) f(u) du = x+3
f(x) -x ∫(0->x) f(u) du +∫(0->x) uf(u) du = x+3
两边取导数
f'(x) -∫(0->x) f(u) du - xf(x) +xf(x) = 1
f'(x) -∫(0->x) f(u) du = 1
两边取导数
f''(x) - f(x) =0
The aux. equation
p^2 -1 =0
p=1 or -1
let
f(x)= Ae^x +Be^(-x)
f(0) =3
A+B=3 (1)
f'(0)=1
A-B = 1 (2)
(1)+(2)
A= 2
from (1)
B= 1
ie
f(x) = 2e^x +e^(-x)