Answer to Question #153982 in Quantitative Methods for usman

Question

y is a function of x satisfying the equation xy″ + ay′ + (x – b) y = 0, where a and b are integers. Find the values of constants a and b if y is given by the following table:

x: 0.8 1 1.2 1.4 1.6 1.8 2 2.2

y: 1.73036 1.95532 2.19756 2.45693 2.73309 3.02549 2.3333 3.65563

Answer 1

Here, $y$ is a function of $x$ satisfying the equation $xy″ + ay' + (x – b) y = 0,$ where $a$ and $b$ are integers.$\def\arraystretch{1.5} \begin{array}{c:c} x &0.8 & 1 &1.2&1.4&1.6&1.8&2&2.2\\ y&1.73036&1.95532&2.19756&2.45693&2.73309&3.02549&2.3333&3.65563\\ \end{array}$

$\text{Now, }\Delta^nf(x_0)=\dfrac{\Delta^{n-1}f(x_1)-\Delta^{n-1}f(x_0)}{x_1-x_0}\\\;\\ \text{and, } \Delta f(x_0)=\dfrac{f(x_1)-f(x_0)}{x_1-x_0}, \Delta f(x_1)=\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\\\;\\ \therefore \Delta f(x_0)=1.1248\\ \Delta^2 f(x_0)=0.216\\ \Delta^3 f(x_0)=-0.003125\\ \Delta^4 f(x_0)=-0.00494792\\ \Delta^5 f(x_0)=-0.000520833\\ \Delta^6 f(x_0)=-21.7025\\ \Delta^7 f(x_0)=108.512$

So,

$f(x)\\=108.512 (x-2) (x-1.8) (x-1.6) (x-1.4) (x-1.2) (x-1) (x-0.8)\\-21.7025 (x-1.8) (x-1.6) (x-1.4) (x-1.2) (x-1) (x-0.8)\\-0.000520833 (x-1.6) (x-1.4) (x-1.2) (x-1) (x-0.8)\\-0.00494792 (x-1.4) (x-1.2) (x-1) (x-0.8)\\-0.003125 (x-1.2) (x-1) (x-0.8)\\+0.216 (x-1) (x-0.8)+1.1248 (x-0.8)+1.73036\\\;\\ = 108.512 x^7-1085.12 x^6+4574.87 x^5-10538.7 x^4+14321.7 x^3-11477.1 x^2+5020.82 x-923.044$

$\text{Now, }\\ y'=f'(x)=759.512x^6-6510.72x^5+22874.35x^4-42154.8x^3+42965.1x^2-22954.2x+5020.82\\ y''=f''(x)=4557.12x^5-32553.6x^4+91497.4x^3-126464.4x^2+85930.2x-22954.2\\$

We have , $xy″ + ay' + (x – b) y = 0$

$\text{Putting }x=1,\\ y'(1)=0.168\\ y''(1)=12.9329\\ \therefore 0.168a-1.95532b+14.88822=0~~~~~~~(1)\\\;\\ \text{Putting }x=0.8,\\ y'(0.8)=6.9152\\ y''(0.8)=-141.1029\\ \therefore 6.9152a-1.73036b=111.15196~~~~~~~~~~(2)$

Solving $(1)\;\text{and }(2),$

We get $a=8,\;b=6$

So, $\pmb{\boxed{ a=8,\;b=6}}$