# The value J ′ ( 0 ) .

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 3.5, Problem 55E

(a)

To determine

## To find: The value J′(0).

Expert Solution

The value J(0)=0.

### Explanation of Solution

Given:

The Bessel function of order 0, y=J(x) is satisfy the differentiable equation xy+y+xy=0 and J(0)=1.

Proof:

Obtain the value J(0).

Since y=J(x) is satisfy the differential equation xy+y+xy=0,

xJ(x)+J(x)+xJ(x)=0

Substitute x=0,

(0)J(0)+J(0)+(0)J(0)=0J(0)=0

Therefore, the value J(0)=0.

(b)

To determine

### To find: The value J′′(0).

Expert Solution

The value J(0)=12.

### Explanation of Solution

Calculation:

Obtain the value J(0).

Since y=J(x) is satisfy the differential equation xy+y+xy=0,

xJ(x)+J(x)+xJ(x)=0x[J(x)+J(x)]+J(x)=0

Differentiate implicitly with respect to x,

ddx(x[J(x)+J(x)]+J(x))=0ddx(x[J(x)+J(x)])+ddx(J(x))=0ddx(x[J(x)+J(x)])+J(x)=0

Apply the product rule and simplify the terms,

xddx([J(x)+J(x)])+[J(x)+J(x)]ddx(x)+J(x)=0x[J(x)+J(x)]+[J(x)+J(x)]+J(x)=0

Substitute x=0,

0[J(0)+J(0)]+[J(0)+J(0)]+J(0)=00+J(0)+J(0)+J(0)=02J(0)+J(0)=0

Substitute J(0)=1,

2J(0)+1=02J(0)=1J(0)=12

Therefore, the value J(0)=12.

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