Q. Approximate the solutions of y' = 3x² , y(1) = 1, using Taylor Series method, using first four terms approximate the value of y(2).

2080 Bsc.CSIT

Solution:

The given differential equation is \( y' = 3x^2 \).

The Taylor series expansion for \( y(x) \) around \( x = 1 \) is given by:

\(y(x) = y(1) + y'(1) x + \frac{y''(1)}{2!} x^2 + \frac{y'''(1)}{3!} x^3 + \ldots\)


The first three derivative of y(x) with respect to x is :

\(y' = 3x^2\)

\(y'' = 6x\)

\(y''' = 6\)

Now, evaluating these at given point x = 1 ,

\(y' = 3x^2 = 3 . 1^2 = 3 \)

\(y'' = 6x = 6.1 = 6\)

\(y''' = 6 \)


By taylors series, 

\(y(x) = y(a) + y'(a)(x-a) + \frac{y''(a)}{2!}(x-a)^2 + \frac{y'''(a)}{3!}(x-a)^3 + \ldots\)

\(= y(1) + y'(1)(x-1) + \frac{y''(1)}{2!}(x-1)^2 + \frac{y'''(1)}{3!}(x-1)^3 + \ldots\)

\(= 1 + 3.(x-1) + \frac{6}{2!}(x-1)^2 + \frac{6}{3!}(x-1)^3 + \ldots\)

\(= 1 + 3x-3 + 3(x^2 -2x+1) + (x^3 -3x^2 + 3x -1) \)

\(= 1 + 3x-3 + 3x^2 -6x+3 + x^3 -3x^2 + 3x -1 \)

∴ \(y(x) = x^3\) which is the required equation.

Now, 

\(y(2) = 2^3 = 8 \)

 

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