Q24E
Question
In Problems 21-28, use the procedure illustrated in Problem 20 to find at least the first four nonzero terms in a power series expansion abouts x=0 of a general solution to the given differential equation.
y"2xy'+3y=x2
Step-by-Step Solution
VerifiedThe first four non-zero terms in a power series expansion to the given differential equation y"2xy'+3y=x2 are y(x)=a0 (1-3x2/2-x4/8+...)+a1 (x-x3/6+...) + (x4/12+...).
The power series approach is used in mathematics to find a power series solution to certain differential equations. In general, such a solution starts with an unknown power series and then plugs that solution into the differential equation to obtain a recurrence relation for the coefficients.
A differential equation's power series solution is a function with an infinite number of terms, each holding a different power of the dependent variable.
It is generally given by the formula,
y(x)=Σn=0∞ anxn
Given,
y"2xy'+3y=x2
Use the formula,
y(x)=Σn=0∞ anxn
Taking derivative of the above equation,
y'(x)=Σn=1∞ nanxn-1
y"(x)=Σn=2∞ n(n-1)anxn-2
Substitute the values in the above formula you get,
Σn=2∞ n(n-1)anxn-2 -2x Σn=1∞ nanxn-1 + 3Σn=0∞ anxn =x2
In order to make the exponent common for all terms you will make the substitution n-2=k therefore k=n+2 in first term and k=n+2 in the other term.
Σk=0∞ (k+2)(k+1)ak+2 xk-2Σk=1∞ k ak xk+3 Σk=0∞ ak xk =x2
2a2+3a0+Σk=1∞ [(k+2)(k+1)ak+2+(3-2k)ak] xk =x2
Hence the expression is 2a2+3a0+Σk=1∞ [(k+2)(k+1)ak+2+(3-2k)ak] xk =x2.
Expand the expression given in the previous step.
(2a2+3a0)+(6a3+a1) x+(12a4-a2)x2+.....=x2
By equating the coefficients you get,
2a2+3a0=0
a2= -3a0/2
6a3+a1=0
a3=-a1/6
12a4-a2=1
a4=1/12-a0/8
Substitute the coefficient.
y(x)=a0+a1x+((-3a0)/2)x2+((-a1)/6)x3+(1/12-a0/8) x4+...
=a0 (1-3x2/2-x4/8+...)+a1 (x-x3/6+...) + (x4/12+...)
Hence, the first four non-zero terms is y(x)=a0 (1-3x2/2-x4/8+...)+a1 (x-x3/6+...) + (x4/12+...).