find power series solution of the differential equation xy"+y'+xy=0

Question

plz its urgent give me the solution

Answer 1

x•y"+y'+xy=0 Let y=summation power +∞ base n=0 a base n X^n,

Answer 2

Let y=summation power+∞ base n=0 a base n X^n, y'= summation power +∞ base n=1 na base n X^n-1, y"=summation power +∞ base n=2 n(n-1)a base nX^n-2 Then, the given becomes: summation +∞ base n=2 n(n-1)a base n X^n-1 + summation +∞ base n=0 a base n•X^n+1 + summation +∞ base n=1 na base n•X^n-1 =0 , Summation power +∞ base n=0 (n+1)•(n+2) a base n+2 X^n+1 + summation power +∞ base n=0 a base n•X^n+1 = - summation power +∞ base n=0 • a base n •X^n+1 , The reccurence formula is :a base n+2= -a base n/(n+2)^2 a base o _= a base o a base 1 _= a base 1 a base 2= - a base o/2^2 a base 3= - a base 1/3^2 a base 4= + a base o/(2•4)^2 a base 5= + a base 1/(3•5)^2 a base 6= - a base o/(2•4•6)^2 a base 7= - a base 1/(3•5•7)^2 Therefore: y, ≈ a base o •[1-(x^2/2)^1 + (x^2/2•4)^2 - (x^2/2•4•6)^3 + — ....] , y base 2 ≈ a base 1 •[x - x^3/3^2 + x^5/(3•5)^2 - x^7/(3•5•7)^2 + — ...] (R=∞) Working out the pattern we ibtain: y(x)= a base o • summation power +∞ base n=0 (-1)^n • X^2n/2^2n • (n!)^2 + a base 1 • summation power +∞ base n=0 (-1)^n • 2^2n • (n!)^2/[(2n+1)!]^2 X^2n+1

Answer 3

Let F=xy"+y'+xy=0.

A power series for x can be written:

y=a0+a1x+a2x^2+a3x^3+a4x^4+...+a[r]x^r+...; or y=∑a[r]x^r for 0≤r≤∞ or y=∑a[r]x^r{0,∞}.

y'=a1+2a2x+3a3x^2+4a4x^3+...+ra[r]x^(r-1)+...; or y'=∑ra[r]x^(r-1){0,∞};

y"=2a2+6a3x+12a4x^2+...+r(r-1)a[r]x^(r-2)+...; or y"=∑r(r-1)a[r]x^(r-2){0,∞};

Substituting for y and its derivatives in F:

xy"=2a2x+6a3x^2+12a4x^3+...+r(r-1)a[r]x^(r-1)+...; or xy"=∑r(r-1)a[r]x^(r-1){0,∞};

y'=a1+2a2x+3a3x^2+4a4x^3+...+ra[r]x^(r-1)+...; or y'=∑ra[r]x^(r-1){0,∞};

xy=a0x+a1x^2+a2x^3+a3x^4+a4x^5+...+a[r]x^(r+1)+...; or y=∑a[r]x^r(r+1) for 0≤r≤∞ or y=∑a[r]x^r{0,∞}.

We can add xy" to y' combining like powers of x: xy"+y'=∑(r(r-1)a[r]x^(r-1)+ra[r]x^(r-1)){0,∞}.

This simplifies because ra[r]x^(r-1) term cancels out and we get xy"+y'=∑r^2a[r]x^(r-1){0,∞}.

To add in the last term xy so that the same power of x is used we have to step back two terms. Let's see what happens if we expand the summation:

xy"+y'+xy=a1+x(a0+4a2)+x^2(a1+9a3)+x^3(a2+16a4)+x^4(a3+25a5)+...

A pattern emerges so we can write F=a1+∑x^(r-1)(a[r-2]+r^2a[r]{2,∞}=0.

The terms have to sum to zero, so a1=0, implying that a3, a5, etc.=0.

Now we need to find a relationship between a0, a2, a4, etc.

a[r-2]+r^2a[r]=0, so a[r]=-a[r-2]/r^2. When r=2, we get a2=-a0/2^2.

From this it follows that, putting r=4, we get a4=-a2/4^2=(-1/4^2)(-1/2^2)a0=a0/((2^2)(4^2)).

And when r=6: a6=-a0/((2^2)(4^2)(6^2)). a8=a0/((2^2)(4^2)(6^2)(8^2)). We can see a pattern emerging.

(2^2)(4^2)(6^2)(8^2)=(2*4*6*8)^2=(1*2*3*4)^2 * (2^2)^4. If r=2n, then in general we get ((n!)^2)(2^2n).

[CHECK: n=1, (1^2)(2^2)=4; n=2, (2^2)(2^4)=64; n=3, (6^2)(2^6)=2304.]

But terms have alternating signs. When n=1, the sign is negative, when n=2 it's positive so we introduce the factor (-1)^n to get the right sign. Write a0 as constant a, and then all other terms are multiples of a (effectively a constant of integration).

For example, a2=-a/4, a4=-a2/16=a/64, a6=-a4/36=-a/2304, etc.

Therefore y=a(1-x^2/4+x^4/64-x^6/2304+...+(-1)^n.x^(2n)/((n!)^2)(2^2n))+...).