Question

use taylors series method to find your at x=0.1 0.2 0.3 considering terms up to the third degree given that dy/dx=x^2+y^2 and y (0)=1 ​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

TAYLOR SERIES METHOD

$\sf{ \: y(x) = y(0) + x {y} \: {'}(0) + \frac{ {x}^{2} }{2! } {y} {''}(0) +\frac{ {x}^{2} }{3! } {y}{'''}(0) + ...... \: }$

gives the values of y for every value of x for which above equation converges

GIVEN

$\sf{{y} \: {'} = {x}^{2} + {y}^{2} } \: \: and \: \: y(0) = 1$

TO DETERMINE

Using Taylor Series Method to find value of y at x=0.1 , 0.2 , 0.3 considering terms up to the third degree

EVALUATION

Here

$\sf{{y} \: {'} = {x}^{2} + {y}^{2} }$

Differentiating both sides with respect to x two times

${y} \: {''} = 2x + 2y{y} \: {'}$

${y} \: {'''} = 2 + 2y{y} \: {''} + 2 {( \: {y} \: {'}\: )}^{2}$

Now it is given that

$y(0) = 1$

So From above

${y} \: {'}(0) = 1$

${y} \: {''} = 0 +( 2 \times 1 \times 1) = 2$

${y} \: {'''} (0)= 2 + 4 + 2 = 8$

So

$\sf{ \: y(x) = y(0) + x {y} \: {'}(0) + \frac{ {x}^{2} }{2! } {y} {''}(0) +\frac{ {x}^{2} }{3! } {y}{'''}(0) + ...... \: }$

$\implies \: \sf{ \: y(x) = 1+ x + \frac{ {x}^{2} }{2! } \times 2 +\frac{ {x}^{2} }{3! } \times 8 + ...... \: }$

$\displaystyle \: \implies \: \sf{ \: y(x) = 1+ x + {x}^{2} + \frac{4 {x}^{3} }{3} \: \: } + ........$

Hence

$\displaystyle \: \sf{ \: y(0.1) = 1 + 0.1 + 0.01 + \frac{4}{3} \times 0.001 \: } = 1.111$

$\displaystyle \: \sf{ \: y(0.2) = 1 + 0.2 + 0.04 + \frac{4}{3} \times 0.008 \: } = 1.251$

$\displaystyle \: \sf{ \: y(0.3) = 1 + 0.3 + 0.09 + \frac{4}{3} \times 0.027 \: } = 1.426$