Q28E
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 about’s x=0 of a general solution to the given differential equation.
y"-(sin x)y=cos x
Step-by-Step Solution
VerifiedThe first four nonzero terms in a power series expansion to the given differential equation y"-(sin x)y=cos x are y(x)=a0+a1x+1/2 x2+a0/6 x3 + (a1/12-1/24) x4+... .
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"-(sin x)y=cos x
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
The series expansion is,
sin x=x-x3/3!+x5/5!-...
cos x=1-x2/2!+x4/4!-...
Substitute the values in the above formula you get,
Σn=2∞ n(n-1)anxn-2 -(x-x3/3!+x5/5!-...) Σn=0∞ anxn
=1-x2/2!+x4/4!-...
(2a2+6a3x+12a4x2+20a5x3+...) - (x-x3/3!+x5/5!-...) (a0+a1x+a2x2+a3x3+...) 1-x2/2!+x4/4!-...
Hence, the expression is (2a2+6a3x+12a4x2+20a5x3+...) - (x-x3/3!+x5/5!-...) (a0+a1x+a2x2+a3x3+...) 1-x2/2!+x4/4!-... .
Expand the expression given in the previous step.
(2a2+6a3x+12a4x2+20a5x3+...) - (a0x2+a1x+a2x3+a3x4+...) + (a0x3/3!+a1x4/3!+a2x5/3!+a3x6/3!+...) - (a0x5/5!+a1x6/5!+a2x7/5!+a3x8/5!+...) + ... =1-x2/2!+x4/4!+...
2a2+(6a3-a0) x+(12a4-a1) x2+...= 1-x2/2!+x4/4!+...
By equating the coefficients you get,
2a2=1
a2=1/2
6a3-a0=0
a3=a0/6
12a4-a1= -1/2
a4=a1/12-1/24
Substitute the coefficient.
y(x)=a0+a1x+1/2 x2+a0/6 x3 + (a1/12-1/24) x4+...
Hence, the first four nonzero terms are y(x)=a0+a1x+1/2 x2+a0/6 x3 + (a1/12-1/24) x4+... .