Approximate the solution of y' = 3x2, y(1) = 1 using Taylor's series m | Numerical Method 2080

Numerical Method20805 marksNumerical: Taylor's series method ODETaylor series method

Approximate the solution of $y' = 3x^2$ , $y(1) = 1$ using Taylor's series method using first four terms. Approximate the value of $y(2)$ .

$y' = 3x^2, \quad y(1) = 1$

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Approximate the solution of $y' = 3x^2$, $y(1) = 1$ using Taylor's series method using first four terms. Approximate the value of $y(2)$.

$$y' = 3x^2, \quad y(1) = 1$$

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ODE: $y' = 3x^2$ - Initial condition: $y(1) = 1$ , so $x0 = 1$ , $y0 = 1$ - Target: $y(2)$ using first four terms of Taylor's series - Step: $h = x - x0 = 2 - 1 = 1$ $y(x) = y(x0) + (x-x0)\frac{y'(x0)}{1!} + (x-x0)^2\frac{y''(x0)}{2!} + (x-x0)^3\frac{y'''(x0)}{3!} + \cdots$ $y' = 3x^2$ $$y'' ...

ODE: $y' = 3x^2$ - Initial condition: $y(1) = 1$, so $x0 = 1$, $y0 = 1$ - Target: $y(2)$ using first four terms of Taylor's series - Step: $h = x - x0 = 2 - 1 = 1$ $$y(x) = y(x0) + (x-x0)\frac{y'(x0)}{1!} + (x-x0)^2\frac{y''(x0)}{2!} + (x-x0)^3\frac{y'''(x0)}{3!} + \cdots$$ $$y' = 3x^2$$ $$y'' ...

Approximate the solution of y' = 3x2, y(1) = 1 using Taylor's series method using first four terms. Approximate the value of y(2).