Solve the given differential equation by means of a power series about the given point x0. Find the first four nonzero terms in each of two linearly independent solutions (unless the series terminates sooner). If possible, find the general term in each solution.

Answered question

2021-11-02

Solve the given differential equation by means of a power series about the given point x0. Find the first four nonzero terms in each of two linearly independent solutions (unless the series terminates sooner). If possible, find the general term in each solution.

(a)yxyy=0,x0=0

(b)yxyy=0,x0=1

(c)(2+4x2x2)y12(x1)y12y=0,x0=1

(d)exy+ln(1+x)yxy=0,x0=0

Answer & Explanation

RizerMix

RizerMix

Expert2023-04-20Added 656 answers

To solve the differential equation y''-xy'-y=0 about the point x0=0, we can assume that the solution can be written as a power series:

y(x)=n=0anxn

where an are constants to be determined.

We can differentiate y(x) with respect to x:

y'(x)=n=1nanxn-1

y''(x)=n=2n(n-1)anxn-2

Substituting these expressions into the differential equation and simplifying, we get:

n=2n(n-1)anxn-2-xn=1nanxn-1-n=0anxn=0

Grouping terms with the same power of x, we get:

a0-a1x+n=2[(n(n-1)an-an-2)xn-nanxn]=0

To satisfy this equation for all values of x, each coefficient of xn must be zero. We can use this to recursively solve for the coefficients an:

n(n-1)an-an-2-nan=0

an=an-2n(n-1)-n

We also have the initial conditions y(0)=a0 and y'(0)=a1. Since y(x) is a power series, we can read off the first few nonzero terms:

a0=y(0)=y0

a1=y'(0)=y'0

a2=a02(2-1)=y02

a3=a23(3-1)-3=-y012-y'06

a4=a24(4-1)-4=y08

Therefore, two linearly independent solutions are:

y1(x)=y0(1+x22+x48+...)

y2(x)=y0x+y'0(1-x26-x424-...)

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