# Power series solutions

```Using Maple to find a power series solution of a differential equation.
As you have seen, solving a differential equation using power series can be tedious. Now
when the purpose is to solve a problem and a suitable device is available, it is often useful
to know how to use the device.
You can do your own searches, if and when you have time, here are a few examples to
get you started.
dy
 y say, using power series you punch in the following commands:
To solve
dx
> with(powseries):
> a:=powsolve(diff(y(x),x)=y(x)):
tpsform(a, x,8);
On hitting enter you would get:
1
1
1
1
1
1
C0C0 x C0 x 2 C0 x 3 C0 x 4
C0 x 5
C0 x 6
C0 x 7O( x 8 )
2
6
24
120
720
5040
Here C0 is a constant and the above line means that the solution is
1
1
1
y  C 0(1  x  12 x 2  16 x 3  241 x 4  120
x 5  720
x 6  5040
x 7  O( x 8 )) (Here O( x 8 ) means
1
x 7 there are terms that involve power 8 or higher of x.
that after 5040
If you do not give any number you would get the series truncated at the fifth power. Here
is the example:
> a:=powsolve(diff(y(x),x)=y(x)):
tpsform(a, x);
1
1
1
1
C0C0 x C0 x 2 C0 x 3 C0 x 4
C0 x 5O( x 6 )
2
6
24
120
Before I let you play with it, I ask you to remember one thing. The software will not give
you a general term of the power series. For example the power series solution given
dy
above does not give you the information that the solution of
dx
y  C 0e x .
Now here is a somewhat nontrivial example: (Recall that if you already have
>with(powseries): on your screen you need not repeat it.)
> a:=powsolve(diff(y(x),x\$2)-y(x)=0):
> tpsform(a,x,8);
1
1
1
1
1
1
C0C1 x C0 x 2 C1 x 3 C0 x 4
C1 x 5
C0 x 6
C1 x 7O( x 8 )
2
6
24
120
720
5040
Here there are two constants involved that means that there are tow linearly independent
solutions. Once you sift though you would find that the solutions are:
1
1
1
y1  1  12 x 2  241 x 4  720
x 6  ... and y 2  x  16 x 3  120
x 5  5040
x 7  ...
Now the question is when you are faced with a situation like:
Find a series solution of y //  xy /  y  0, about x0  1 .. In such a case you can use the zsubstitution: Put z=x-1. This gives you x=z+1. You can change the equation to
y //  ( z  1) y /  y  0, about z 0  0 ,
> a:=powsolve(diff(y(z),z\$2)-(z+1)*diff(y(z),z)-y(z)=0):
> tpsform(a,z,8);
1
1
1
1
1
1
3
1
C0C1 z C1 C0  z2 C1 C0  z3 C1 C0  z4 C1 C0 
2
2
2
6
4
6
20
15








1
7
13
19
z5 C1
C0  z6
C1
C0  z7O( z8 )
15
180
420
1260




Now sift through and rewrite the two solution by substituting z=x-1.
There is the last job of taking care of problems like
y //  xy /  y  0, x0  0, y (0)  2, y / (0)  1
For such problems the command is:
> a:=powsolve(diff(y(x),x\$2)-x*diff(y(x),x)y(x)=0,y(0)=2,D(y)(0)=1):
> tpsform(a,x,8);
1
1
1
1
1 7
2xx 2 x 3 x 4 x 5 x 6
x O( x 8 )
3
4
15
24
105
```
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