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logo1 Overview An Example Double Check

Series Solutions

Bernd Schr¨

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Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

What are Series Solutions?

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

What are Series Solutions?

• 1. A series solution to a differential equation is a solution of

the form y =

∞

∑

n=0

cn(x−x0)n.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

What are Series Solutions?

• 1. A series solution to a differential equation is a solution of

the form y =

∞

∑

n=0

cn(x−x0)n. That is, any solution that can be expanded into a Taylor series is a series solution.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

What are Series Solutions?

• 1. A series solution to a differential equation is a solution of

the form y =

∞

∑

n=0

cn(x−x0)n. That is, any solution that can be expanded into a Taylor series is a series solution.

• 2. For differential equations of the form

y′′ +P(x)y′ +Q(x)y = 0, there are mild technical conditions (P and Q must be analytic at x0) that guarantee that there are series solutions.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

What are Series Solutions?

• 1. A series solution to a differential equation is a solution of

the form y =

∞

∑

n=0

cn(x−x0)n. That is, any solution that can be expanded into a Taylor series is a series solution.

• 2. For differential equations of the form

y′′ +P(x)y′ +Q(x)y = 0, there are mild technical conditions (P and Q must be analytic at x0) that guarantee that there are series solutions.

• 3. If there is a series solution, then we can substitute the

series and its derivatives into the equation to obtain a set of equations for the cn.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

What are Series Solutions?

• 1. A series solution to a differential equation is a solution of

the form y =

∞

∑

n=0

cn(x−x0)n. That is, any solution that can be expanded into a Taylor series is a series solution.

• 2. For differential equations of the form

y′′ +P(x)y′ +Q(x)y = 0, there are mild technical conditions (P and Q must be analytic at x0) that guarantee that there are series solutions.

• 3. If there is a series solution, then we can substitute the

series and its derivatives into the equation to obtain a set of equations for the cn.

• 4. These equations will allow us to compute the cn.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

What are Series Solutions?

• 1. A series solution to a differential equation is a solution of

the form y =

∞

∑

n=0

cn(x−x0)n. That is, any solution that can be expanded into a Taylor series is a series solution.

• 2. For differential equations of the form

y′′ +P(x)y′ +Q(x)y = 0, there are mild technical conditions (P and Q must be analytic at x0) that guarantee that there are series solutions.

• 3. If there is a series solution, then we can substitute the

series and its derivatives into the equation to obtain a set of equations for the cn.

• 4. These equations will allow us to compute the cn.

That’s it.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

What are Series Solutions?

• 1. A series solution to a differential equation is a solution of

the form y =

∞

∑

n=0

cn(x−x0)n. That is, any solution that can be expanded into a Taylor series is a series solution.

• 2. For differential equations of the form

y′′ +P(x)y′ +Q(x)y = 0, there are mild technical conditions (P and Q must be analytic at x0) that guarantee that there are series solutions.

• 3. If there is a series solution, then we can substitute the

series and its derivatives into the equation to obtain a set of equations for the cn.

• 4. These equations will allow us to compute the cn.

That’s it. (Except convergence analysis. Separate topic.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

• ∞

∑

n=0

cnxn ′′ −x

• ∞

∑

n=0

cnxn ′ +x

• ∞

∑

n=0

cnxn

• =

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

• ∞

∑

n=0

cnxn ′′ −x

• ∞

∑

n=0

cnxn ′ +x

∞

∑

n=0

cnxn =

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

• ∞

∑

n=0

cnxn ′′ −x

∞

∑

n=1

cnnxn−1 +x

∞

∑

n=0

cnxn =

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

∞

∑

n=2

cnn(n−1)xn−2 −x

∞

∑

n=1

cnnxn−1 +x

∞

∑

n=0

cnxn =

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

∞

∑

n=2

cnn(n−1)xn−2 −x

∞

∑

n=1

cnnxn−1 +x

∞

∑

n=0

cnxn =

∞

∑

n=2

n(n−1)cnxn−2 −

∞

∑

n=1

ncnxn +

∞

∑

n=0

cnxn+1 =

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

∞

∑

n=2

cnn(n−1)xn−2 −x

∞

∑

n=1

cnnxn−1 +x

∞

∑

n=0

cnxn =

∞

∑

n=2

n(n−1)cnxn−2 −

∞

∑

n=1

ncnxn +

∞

∑

n=0

cnxn+1 = k := n−2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

∞

∑

n=2

cnn(n−1)xn−2 −x

∞

∑

n=1

cnnxn−1 +x

∞

∑

n=0

cnxn =

∞

∑

n=2

n(n−1)cnxn−2 −

∞

∑

n=1

ncnxn +

∞

∑

n=0

cnxn+1 = k := n−2 k := n

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

∞

∑

n=2

cnn(n−1)xn−2 −x

∞

∑

n=1

cnnxn−1 +x

∞

∑

n=0

cnxn =

∞

∑

n=2

n(n−1)cnxn−2 −

∞

∑

n=1

ncnxn +

∞

∑

n=0

cnxn+1 = k := n−2 k := n k := n+1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

∞

∑

n=2

cnn(n−1)xn−2 −x

∞

∑

n=1

cnnxn−1 +x

∞

∑

n=0

cnxn =

∞

∑

n=2

n(n−1)cnxn−2 −

∞

∑

n=1

ncnxn +

∞

∑

n=0

cnxn+1 = k := n−2 k := n k := n+1 n = k +2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

∞

∑

n=2

cnn(n−1)xn−2 −x

∞

∑

n=1

cnnxn−1 +x

∞

∑

n=0

cnxn =

∞

∑

n=2

n(n−1)cnxn−2 −

∞

∑

n=1

ncnxn +

∞

∑

n=0

cnxn+1 = k := n−2 k := n k := n+1 n = k +2 n = k

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

∞

∑

n=2

cnn(n−1)xn−2 −x

∞

∑

n=1

cnnxn−1 +x

∞

∑

n=0

cnxn =

∞

∑

n=2

n(n−1)cnxn−2 −

∞

∑

n=1

ncnxn +

∞

∑

n=0

cnxn+1 = k := n−2 k := n k := n+1 n = k +2 n = k n = k −1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

∞

∑

n=2

cnn(n−1)xn−2 −x

∞

∑

n=1

cnnxn−1 +x

∞

∑

n=0

cnxn =

∞

∑

n=2

n(n−1)cnxn−2 −

∞

∑

n=1

ncnxn +

∞

∑

n=0

cnxn+1 = k := n−2 k := n k := n+1 n = k +2 n = k n = k −1

∞

∑

k=0

(k +2)(k +1)ck+2xk

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

∞

∑

n=2

cnn(n−1)xn−2 −x

∞

∑

n=1

cnnxn−1 +x

∞

∑

n=0

cnxn =

∞

∑

n=2

n(n−1)cnxn−2 −

∞

∑

n=1

ncnxn +

∞

∑

n=0

cnxn+1 = k := n−2 k := n k := n+1 n = k +2 n = k n = k −1

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

y′′ −xy′ +xy =

∞

∑

n=2

cnn(n−1)xn−2 −x

∞

∑

n=1

cnnxn−1 +x

∞

∑

n=0

cnxn =

∞

∑

n=2

n(n−1)cnxn−2 −

∞

∑

n=1

ncnxn +

∞

∑

n=0

cnxn+1 = k := n−2 k := n k := n+1 n = k +2 n = k n = k −1

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk +

∞

∑

k=1

ck−1xk =

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk +

∞

∑

k=1

ck−1xk =

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk +

∞

∑

k=1

ck−1xk = 2c2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk +

∞

∑

k=1

ck−1xk = 2c2 +

∞

∑

k=1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk +

∞

∑

k=1

ck−1xk = 2c2 +

∞

∑

k=1

xk

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk +

∞

∑

k=1

ck−1xk = 2c2 +

∞

∑

k=1

xk (k +2)(k +1)ck+2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk +

∞

∑

k=1

ck−1xk = 2c2 +

∞

∑

k=1

xk (k +2)(k +1)ck+2 −kck

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk +

∞

∑

k=1

ck−1xk = 2c2 +

∞

∑

k=1

xk (k +2)(k +1)ck+2 −kck +ck−1

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk +

∞

∑

k=1

ck−1xk = 2c2 +

∞

∑

k=1

xk (k +2)(k +1)ck+2 −kck +ck−1

• =

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk +

∞

∑

k=1

ck−1xk = 2c2 +

∞

∑

k=1

xk (k +2)(k +1)ck+2 −kck +ck−1

• =

2c2 = 0

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk +

∞

∑

k=1

ck−1xk = 2c2 +

∞

∑

k=1

xk (k +2)(k +1)ck+2 −kck +ck−1

• =

2c2 = 0, c2 = 0

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk +

∞

∑

k=1

ck−1xk = 2c2 +

∞

∑

k=1

xk (k +2)(k +1)ck+2 −kck +ck−1

• =

2c2 = 0, c2 = 0 (k +2)(k +1)ck+2 −kck +ck−1 =

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk +

∞

∑

k=1

ck−1xk = 2c2 +

∞

∑

k=1

xk (k +2)(k +1)ck+2 −kck +ck−1

• =

2c2 = 0, c2 = 0 (k +2)(k +1)ck+2 −kck +ck−1 = (k +2)(k +1)ck+2 = kck −ck−1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

∞

∑

k=0

(k +2)(k +1)ck+2xk −

∞

∑

k=1

kckxk +

∞

∑

k=1

ck−1xk = 2c2 +

∞

∑

k=1

xk (k +2)(k +1)ck+2 −kck +ck−1

• =

2c2 = 0, c2 = 0 (k +2)(k +1)ck+2 −kck +ck−1 = (k +2)(k +1)ck+2 = kck −ck−1 ck+2 = kck −ck−1 (k +2)(k +1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

c0 = y(0) = 0,

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

c0 = y(0) = 0, c1 = y′(0) = 1,

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

c0 = y(0) = 0, c1 = y′(0) = 1, c2 = 0,

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

c0 = y(0) = 0, c1 = y′(0) = 1, c2 = 0, ck+2 = kck −ck−1 (k +2)(k +1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

c0 = y(0) = 0, c1 = y′(0) = 1, c2 = 0, ck+2 = kck −ck−1 (k +2)(k +1) (k = 1) c3 = c1+2 = 1·c1 −c0 (1+2)(1+1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

c0 = y(0) = 0, c1 = y′(0) = 1, c2 = 0, ck+2 = kck −ck−1 (k +2)(k +1) (k = 1) c3 = c1+2 = 1·c1 −c0 (1+2)(1+1) = 1 6

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

c0 = y(0) = 0, c1 = y′(0) = 1, c2 = 0, ck+2 = kck −ck−1 (k +2)(k +1) (k = 1) c3 = c1+2 = 1·c1 −c0 (1+2)(1+1) = 1 6 (k = 2) c4 = c2+2 = 2·c2 −c1 (2+2)(2+1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

c0 = y(0) = 0, c1 = y′(0) = 1, c2 = 0, ck+2 = kck −ck−1 (k +2)(k +1) (k = 1) c3 = c1+2 = 1·c1 −c0 (1+2)(1+1) = 1 6 (k = 2) c4 = c2+2 = 2·c2 −c1 (2+2)(2+1) = − 1 12

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

c0 = y(0) = 0, c1 = y′(0) = 1, c2 = 0, ck+2 = kck −ck−1 (k +2)(k +1) (k = 1) c3 = c1+2 = 1·c1 −c0 (1+2)(1+1) = 1 6 (k = 2) c4 = c2+2 = 2·c2 −c1 (2+2)(2+1) = − 1 12 (k = 3) c5 = c2+3 = 3·c3 −c2 (3+2)(3+1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

c0 = y(0) = 0, c1 = y′(0) = 1, c2 = 0, ck+2 = kck −ck−1 (k +2)(k +1) (k = 1) c3 = c1+2 = 1·c1 −c0 (1+2)(1+1) = 1 6 (k = 2) c4 = c2+2 = 2·c2 −c1 (2+2)(2+1) = − 1 12 (k = 3) c5 = c2+3 = 3·c3 −c2 (3+2)(3+1) = 3· 1

6 −0

20

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

c0 = y(0) = 0, c1 = y′(0) = 1, c2 = 0, ck+2 = kck −ck−1 (k +2)(k +1) (k = 1) c3 = c1+2 = 1·c1 −c0 (1+2)(1+1) = 1 6 (k = 2) c4 = c2+2 = 2·c2 −c1 (2+2)(2+1) = − 1 12 (k = 3) c5 = c2+3 = 3·c3 −c2 (3+2)(3+1) = 3· 1

6 −0

20 = 1 40

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

c0 = y(0) = 0, c1 = y′(0) = 1, c2 = 0, ck+2 = kck −ck−1 (k +2)(k +1) (k = 1) c3 = c1+2 = 1·c1 −c0 (1+2)(1+1) = 1 6 (k = 2) c4 = c2+2 = 2·c2 −c1 (2+2)(2+1) = − 1 12 (k = 3) c5 = c2+3 = 3·c3 −c2 (3+2)(3+1) = 3· 1

6 −0

20 = 1 40 (k = 4) c6 = c2+4 = 4·c4 −c3 (4+2)(4+1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

c0 = y(0) = 0, c1 = y′(0) = 1, c2 = 0, ck+2 = kck −ck−1 (k +2)(k +1) (k = 1) c3 = c1+2 = 1·c1 −c0 (1+2)(1+1) = 1 6 (k = 2) c4 = c2+2 = 2·c2 −c1 (2+2)(2+1) = − 1 12 (k = 3) c5 = c2+3 = 3·c3 −c2 (3+2)(3+1) = 3· 1

6 −0

20 = 1 40 (k = 4) c6 = c2+4 = 4·c4 −c3 (4+2)(4+1) = 4·

• − 1

12

• − 1

6

30

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1 (continued)

c0 = y(0) = 0, c1 = y′(0) = 1, c2 = 0, ck+2 = kck −ck−1 (k +2)(k +1) (k = 1) c3 = c1+2 = 1·c1 −c0 (1+2)(1+1) = 1 6 (k = 2) c4 = c2+2 = 2·c2 −c1 (2+2)(2+1) = − 1 12 (k = 3) c5 = c2+3 = 3·c3 −c2 (3+2)(3+1) = 3· 1

6 −0

20 = 1 40 (k = 4) c6 = c2+4 = 4·c4 −c3 (4+2)(4+1) = 4·

• − 1

12

• − 1

6

30 = − 1 60

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

• 1. The series expansion of the solution starts with the

polynomial y(x) = x+ 1 6x3 − 1 12x4 + 1 40x5 − 1 60x6.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

• 1. The series expansion of the solution starts with the

polynomial y(x) = x+ 1 6x3 − 1 12x4 + 1 40x5 − 1 60x6.

• 2. It should be noted that the above polynomial is not the

solution itself.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Solve the Initial Value Problem y′′ −xy′ +xy = 0, y(0) = 0, y′(0) = 1

• 1. The series expansion of the solution starts with the

polynomial y(x) = x+ 1 6x3 − 1 12x4 + 1 40x5 − 1 60x6.

• 2. It should be noted that the above polynomial is not the

solution itself.

• 3. To emphasize this fact, the solution can be stated as

y(x) = x+ 1 6x3 − 1 12x4 + 1 40x5 − 1 60x6 +···.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Checking Series Solutions When Only The First Few Terms are Available

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Checking Series Solutions When Only The First Few Terms are Available

• 1. Checking the initial conditions is easy, because y(x0) = c0

and y′(x0) = c1.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Checking Series Solutions When Only The First Few Terms are Available

• 1. Checking the initial conditions is easy, because y(x0) = c0

and y′(x0) = c1.

• 2. Because the “solution” is a truncated series, we cannot

expect that the differential equation is exactly satisfied.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Checking Series Solutions When Only The First Few Terms are Available

• 1. Checking the initial conditions is easy, because y(x0) = c0

and y′(x0) = c1.

• 2. Because the “solution” is a truncated series, we cannot

expect that the differential equation is exactly satisfied.

• 3. What should happen, though, is that low order terms
• cancel. If your approximation goes up to, say x9, and, after

the check, there is a term with x2 left over, then, most likely, there is a problem.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Checking Series Solutions When Only The First Few Terms are Available

• 1. Checking the initial conditions is easy, because y(x0) = c0

and y′(x0) = c1.

• 2. Because the “solution” is a truncated series, we cannot

expect that the differential equation is exactly satisfied.

• 3. What should happen, though, is that low order terms
• cancel. If your approximation goes up to, say x9, and, after

the check, there is a term with x2 left over, then, most likely, there is a problem.

• 4. Because the computations are quite tedious, use a

computer algebra system.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Checking Series Solutions When Only The First Few Terms are Available

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Checking Series Solutions When Only The First Few Terms are Available

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Checking Series Solutions When Only The First Few Terms are Available

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions

logo1 Overview An Example Double Check

Checking Series Solutions When Only The First Few Terms are Available

(Remember that we went up to k = 4, so terms up to order 4 must cancel.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Series Solutions