Question

解一阶常微方程？

(a)求一阶常微方程 4x dy/dx-4y=9(∛(xy^2 )) 是否齐次并以换元法解开（以y表达x）。(b)找特解，当 y(1)=-125看图会清楚一点

Answer 1

(1)
let
u= y/x
y= xu
dy/dx = x.du/dx + u
//
4xy'-4y=9x^(1/3).y^(2/3)

y'-(y/x)= (9/4)(y/x)^(2/3)
x.du/dx + u - u = (9/4)u^(2/3)
x.du/dx =(9/4)u^(2/3)
∫u^(-2/3) du =(9/4)∫ dx/x
3u^(1/3) = (9/4)lnx + C
3(y/x)^(1/3) = (9/4)lnx + C
y^(1/3) = x^(1/3). [ (9/4)lnx + C ] /3
y= { x^(1/3). [ (9/4)lnx + C ] /3 }^3
= (x/27)[ (9/4)lnx + C ]^3
(2)
y(1) =-125
-125= ( C/3)^3
C/3 = -5

C=-15
ie
y= (x/27)[ (9/4)lnx -15 ]^3