Question

taylor series,find y(1.1) given y^1=x+y, y(1)=0
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Answer 1

Answer:

Taylor's Series method ... f is a function of two variables x and y and (x0 , y0) is a known point on the ... Example 4, Find y at x = 1.1 and 1.2 by solving y' = x2 + y2 , y(1) = 2.3, Solution ...

Answer 2

Answer:

The value of y(1.1) using Taylor’s series expansion is 0.11033.

Step-by-step explanation:

Given the first order equation, $y^{'}=x+y$

and $y(1)=0$ where x = 1, y = 0

Substituting x = 1, y = 0, in the given first order equation,

$y^{'}=x+y \implies y^{'}=1+0=1$

Differentiating the given first order equation with respect to x

$y^{''}=1+y^{'}$

and substituting $y^{' }= 1$

$\implies y^{''}=1+1=2$

Differentiating $y^{''}$ with respect to x again and substituting $y^{'' }= 2$

$y^{'''}=y^{''} \implies y^{'''}=2$

The Taylor's series expansion is given by

$y=y_{0} +(x-x_{0})y_{0}^{'} +\dfrac{(x-x_{0})^{2} }{2!} y_{0}^{''} +\dfrac{(x-x_{0})^{3} }{3!} y_{0}^{'''} +...$

The series gives the values of y for every value of x for which the expansion converges.

Given  $y(1)=0$ , therefore, applying the Taylor’s series about x=1,

$y(x)=y(1) +(x-1)y^{'} (1)+\dfrac{(x-1)^{2} }{2!} y^{''} (1) +\dfrac{(x-1)^{3} }{3!} y^{''} (1) +...$

Substituting the obtained values in the above,

$y(x)=0 +(x-1)\times1+\dfrac{(x-1)^{2} }{2!} \times 2 +\dfrac{(x-1)^{3} }{3!} \times2$

For $y(1.1)$, where $x = 1.1$, we get

$y(1.1)=(1.1-1)\times1+\dfrac{(1.1-1)^{2} }{2!} \times 2 +\dfrac{(1.1-1)^{3} }{3!} \times2$

$=(0.1)+\dfrac{(0.1)^{2} }{2} \times 2 +\dfrac{(0.1)^{3} }{3\times2} \times2$

$=(0.1)+{(0.1)^{2} +\dfrac{(0.1)^{3} }{3}$

$=0.1+0.01+0.00033=0.11033$

Therefore, the value of y(1.1) using Taylor’s series expansion is 0.11033.