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