Question

Euler 's Cauchy equation












Find general solution for
1) 4x2y” + 4xy’-y= 0
2) 2x2y” + 4xy’-5y= 0
Solve IVP for
1) x2y” - 4xy’+6y= 0 , y(1) = 1 ,y’(1) =0
2) x2y” +3xy’+ y= 0 , y(1) = 4 ,y’(1) =-2

Answer 1

\[x^2\frac{\mathrm d^2y}{\mathrm dx^2}+\alpha x\frac{\mathrm dy}{\mathrm dx}+by=f(x)\]

substituting \(x=e^z\)

\[\qquad\Downarrow\qquad\Downarrow\qquad\Downarrow\]
\[\frac{\mathrm d^2y}{\mathrm dz^2}+(\alpha-1)\frac{\mathrm dy}{\mathrm dz}+by=f(e^z)\]

Answer 2

to Find general solution for
1) 4x2y” + 4xy’-y= 0
as in this differential equcation
degree of coefficient=order of derivative
and it fit to form
ax^2y''+bxy'+cy=0 ( so we use cauchy euler equcation)
by comparing both eq we have
a=4 ,b=4 ,c=-1
in cauchy euler equcation we use equcation
am(m-1)+bm+c=0
putting values of a,b and c
4m(m-1)+4m-1=0
4m^2-4m-1=0
4m^2=1
m=1/2 and m=-1/2
as solution for m are real ans distint so general solution is in form
y=c1x^(1/2)+c2x(-1/2)

Answer 3

Solve IVP for
1) x2y” - 4xy’+6y= 0 , y(1) = 1 ,y’(1) =0
a=1 ,b=-4 ,c=6
so
m(m-1)-4m+c=0
m^2-5m+6=0
(m-2)(m-3)=0
m=2,m=3
two different real root of m so general solution
y=c1x^2+c2x^3
now we find particular solution by using iv
as y(1)=1
so y=c1+c2 (1 eq)
as y'(1)=0
by taking derivative of y=c1x^2+c2x^3
y'=2c1x+3c2x^2
0=2c1+2c2 (2 eq)

by solving eq 1 and 2 we got
c1=3
c2=-2
so particular solution

y=3x^2-2x^3