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Verify the formula stated for y0,y1and y2in terms of sinxand cosxand also find the value of the y3/2(x) and y5/2(x).

Short Answer

Expert verified

The values of the y3/2(x),and y5/2(x) are localid="1659330486663" y32(x)=-2Ï€x-32sin(x)+xcos(x)and y52(x)=-2Ï€x-52sin(x)+x2-3xcos(x) , respectively and formula stated for y0,y1

and y2in terms of sin x and cos x are also proved.

Step by step solution

01

 Step 1: Concept of recursion relation:

The recursion relation is given as follows:

ddx(x-pJpx)=-x-1Jp+1(x)

02

Find the value of y3/2(x), and y5/2(x) :

As the given equation is,

Y12(x)=-2Ï€xcos(x)

As you know,

localid="1659330970652" yn(x)=Ï€2x.Y2n+12(x)yn(x)=-xn-1xddxncos(x)x........(1)

For n = 0,1,2...:

ddxx-pJp(x)=-x-pJp+1(x)..........(2)

From equation (2) obtain:

ddxx∑21Y12(x)=x12Y32(x)ddxx32Y32(x)=x-32Y62(x)

Therefore,

Y32(x)=-x12ddxx12-2Ï€³æcos(x)=-x12ddx1x2Ï€³æcos(x)=2Ï€³æ32sin(x)+xcos(x)

03

Find the value of y5/2(x) :

Similarly for Y52(x):

Y52(x)=x32-ddx-2Ï€x-32sin(x)+xcos(x)=x32-ddxx-32Ï€sin(x)+xcos(x)=-2Ï€x-523xsin(x)+(x2-3)cos(x)

From equation (1) obtain:

Y0(x)=2Ï€.Y12(x)=2Ï€2Ï€³æcos(x)=-cos(x)x=-x0-1xddx0cos(x)x

For, n = 0 .

Now,

Y!(x)=2Ï€.Y32(x)=22x-2Ï€x32sin(x)+xcos(x)=-x2sin(x)+xcos(x)=-x1-1xddx1cos(x)x,forn=1

04

Find the value of y2(x) :

Similarly for y2(x):

Y2(x)=Ï€2xY52(x)=Ï€2x-2Ï€x-523xsin(x)+(x2-3)cos(x)=-x-33xsin(x)+(x2-3)cos(x)=x2-1xddx2cos(x)x,forn=2

With the following formula.

-1xddx2sin(x)x=-1xddx-1xddx2cos(x)x

Therefore, you obtain,

Y32(x)=-2Ï€x-32sin(x)+cos(x)Y52(x)=-2Ï€x-523xsin(x)+x2-3cos(x)

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