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1 Answer

How do I solve this problem: y’’.x.ln(x) = y’’

Since you have y’’ on both sides, then you can cancel both instances, giving

x.ln(x) = 1

ln(x) = 1/x,

which we can write as

f(x) = ln(x) – 1/x   (=0)

We can now use the Newton-Raphson method to give an iterative solution.

The iterative sequence is,

x_(n+1) = x_n – f(x_n) / f’(x_n)   ------------------------------ (1)

 

Starting value (x_1 = ?)

At x = 1: f(1) = ln(1) – 1/1 = -1

At x = 2: f(2) = ln(2) – ½ = 0.1931

There is a change of sign, so the root occurs between x = 1 and x = 2.

Let x_1 = 1.

 

Now tabulate the calculations

n

x_n

f(x_n)

f’(x_n)

x_(n+1) using (1)

1

1

-1

2

1.5

2

1.5

-0.2612

1.1111

1.73508

3

1.73508

-0.02528

0.9085

1.76292

4

1.76292

-2.7328x10^(-4)

0.8891

1.76322

5

1.78322

-3.2446x10^(-8)

0.8888

1.76322

 

Solution: x = 1.76322

by Level 11 User (81.5k points)

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