I'm trying to find particular solution of the second-order linear

protisvitfc

protisvitfc

Answered question

2021-11-19

I'm trying to find particular solution of the second-order linear equation but I can't find y1 and y2 according to y=c1y1+c2y2
x2y2xy+2y=0, y(1)=3, y(1)=1
If r is used, x2r22xr=2 then xr(xr2)=2, I can't go on from there to find y1 and y2

Answer & Explanation

Roger Noah

Roger Noah

Beginner2021-11-20Added 17 answers

Assume y=xr then if y=xr then
y=rxr1
And
y=r(r1)xr2
Substitute in
x2y2xy+2y=0
to get
r(r1)xr2rxr+2xr=0
divide by y to get
r(r1)2r+2=(r1)(r2)=0
therefore
y=ax+bx2
Solve
y(1)=a+b=3
y(1)=a+2b=1
to get y=5x2x2
Charles Wee

Charles Wee

Beginner2021-11-21Added 14 answers

y(x)x2+2y(x)x+2y(x)=0
Assume a solution to this Euler-Cauchy equation will be proportional to eλ for some constant λ
Substitute y=xλ into the differential equation:
x2d2dx2(xλ)+2xddx(xλ)+2xλ=0
Substitute d2dx2(xλ)=(λ1)xλ2 and ddx(xλ)=λxλ1
λ2xλ+λxλ+2xλ=0
xλ(λ2+λ+2)=0
Assuming x0, the zeros must come from the polynomial:
λ2+λ+2=0
λ=12±i72
The roots λ=12±i22 give y1(x)=C1x12+i72+C2x12i72 as solutions, where C1 and C2 are arbitrary constants. The general solution is the sum of the above solutions:
y(x)=y1(x)+y2(x)=C1x12+i72+C2x12i72
Using xλ=eλln(x), apply Euler's identity eα+bi=eαcos(b)+ieαsin(b)

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