Question #153982

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


1
Expert's answer
2021-01-11T16:52:36-0500

Here, yy is a function of xx satisfying the equation xy+ay+(xb)y=0,xy″ + ay' + (x – b) y = 0, where aa and bb are integers.x0.811.21.41.61.822.2y1.730361.955322.197562.456932.733093.025492.33333.65563\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}


Now, Δnf(x0)=Δn1f(x1)Δn1f(x0)x1x0  and, Δf(x0)=f(x1)f(x0)x1x0,Δf(x1)=f(x2)f(x1)x2x1  Δf(x0)=1.1248Δ2f(x0)=0.216Δ3f(x0)=0.003125Δ4f(x0)=0.00494792Δ5f(x0)=0.000520833Δ6f(x0)=21.7025Δ7f(x0)=108.512\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(x2)(x1.8)(x1.6)(x1.4)(x1.2)(x1)(x0.8)21.7025(x1.8)(x1.6)(x1.4)(x1.2)(x1)(x0.8)0.000520833(x1.6)(x1.4)(x1.2)(x1)(x0.8)0.00494792(x1.4)(x1.2)(x1)(x0.8)0.003125(x1.2)(x1)(x0.8)+0.216(x1)(x0.8)+1.1248(x0.8)+1.73036  =108.512x71085.12x6+4574.87x510538.7x4+14321.7x311477.1x2+5020.82x923.044f(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

Now, y=f(x)=759.512x66510.72x5+22874.35x442154.8x3+42965.1x222954.2x+5020.82y=f(x)=4557.12x532553.6x4+91497.4x3126464.4x2+85930.2x22954.2\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+(xb)y=0xy″ + ay' + (x – b) y = 0


Putting x=1,y(1)=0.168y(1)=12.93290.168a1.95532b+14.88822=0       (1)  Putting x=0.8,y(0.8)=6.9152y(0.8)=141.10296.9152a1.73036b=111.15196          (2)\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)  and (2),(1)\;\text{and }(2),

We get a=8,  b=6a=8,\;b=6


So, a=8,  b=6\pmb{\boxed{ a=8,\;b=6}}

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