# Series Solution of a Cauchy-Euler Equation

Series Solution of a Cauchy-Euler Equation

′

y

-8

10

′′

y

′

y

′

y

′

y

-8

10

′′

y

′

y

′

y

′

y

′′

y

′

y

′

y

This Demonstration shows the solution to the Cauchy–Euler equation with initial conditions and and approximations to it using truncated series.

2y''(x)-xy'(x)+(1+x)y(x)=0

2

x

y(1)=1

y'(1)=0

Assume solutions have the form

y=+x++…=

r

x

a

0

a

1

a

2

2

x

r

x

∞

∑

n=0

a

n

n

x

Take the first and second derivatives of this equation and substitute back into the original equation. If the equation is to be satisfied for all , the coefficient of each power of must be zero. This gives a quadratic equation in with roots and . Then the coefficients , , , … can be determined. The two solutions for are:

x

x

r

r=1

r=1/2

a

0

a

1

a

2

x>0

y

1

∞

∑

n=1

n

(-1)

n

2

(2n+1)!

n

x

y

2

1/2

x

∞

∑

n=1

n

(-1)

n

2

(2n)!

n

x

The final solution (plotted in blue) has the form , where and are determined by the initial conditions. With not too many terms it serves as a good approximation to the exact solution (plotted in red).

y=(x)+(x)

c

1

y

1

c

2

y

2

c

1

c

2