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Messages - XueQiWang

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Quiz-3 / TUT0801 quiz3
« on: October 14, 2019, 08:00:29 PM »
Q：Find the Wronskian of two solutions of the given differential equation without solving the equation.
x^2·y''+xy'+(x^2-v^2)y=0

A:
x^2·y''+xy'+(x^2-v^2)y=0
both side divided by x^2:  y''+y'/x+(x^2-v^2)y/x^2=0
such that: p(x)=1/x
W=ce^(-∫p(x)dx)=ce^(-∫(1/x)dx)=ce^(-ln(x)+c)=ce^(ln(1/x)+c)=cx^-1·e^c
therefore: W=c/x

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Quiz-2 / TUT0801 quiz2
« on: October 05, 2019, 01:03:22 AM »
Question: x^2·y^3+x(1+y^2)y’=0  μ(x,y)=1/xy^3

Answer:

let x^2·y^3 be M(x,y)  and  x(1+y^2) be N(x,y)
M(y)=d(x^2y^3)/dy=3x^2·y^2
N(x)=d(x(1+y^2))/dx=1+y^2
M(y) is not equal to N(x), which means given equation is not an exact equation.

Next step, multiply both side of M(y) and N(x) with the integrating factor μ(x,y)=1/xy^3
1/xy^3(x^2·y^3+x(1+y^2)y’)=0
x+((1+y^2)/y^3)y’=0

let x be the M1(x,y) and (1+y^2)/y^3 be N1(x,y)
M1(x,y)=d(x)/dy=0
N1(x,y)=d((1+y^2)/y^3)/dx=0
As M1(x,y)=N1(x,y). Therefore, the equation is an exact equation.

∃φ(x,y) s.t. φx=M1
 φ=∫M1dx=∫xdx=1/2 x^2+h(y)
 φy=h(y)'=N1=(1+y^2)/y^3
h(y)=-1/2y^-2+ln|y|
 φ=1/2 x^2-1/2y^-2+ln|y|=C
This is the general solution of this question

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