Question

2
b) Using Taylor's series method find y at x =1.1 and
1.2 by solving dy/dx = x2 + y2 given y(1)=2.3.​

Answer 1

Answer:

Using Taylor’s series method find y at x=1 .1 and 1.2 by solving,

2 2

x y

dx

dy

 

given y(1)=2.3

Answer 2

Answer:

y at 1.1 is approximately 3.26

and y at 1.2 is approximately 3.69

Step-by-step explanation:

Given:

y(1) = 2.3 => x₀ = 1 and y₀ = 2.3 (initial conditions)

$\frac{dy}{dx}$ = x² + y² => y' = x² + y²

Now,

y' = x² + y² => y₀' = 1² + (2.3)² = 6.29

y'' = 2x + 2y => y₀'' = 2*1 + 2*2.3 = 6.6

y''' = 2 + 2 => y₀''' = 4

y'''' = 0 + 0 = 0 = y₀''''

By Taylor's Theorem,

We know that y(x) = y₀ + (x - x₀)y₀' + $\frac{(x-x_0)^2}{2!}$y₀'' +  $\frac{(x-x_0)^3}{3!}$y₀''' + . . . . . . .

=> y(x) = 2.3 + (x-1) * (6.29) +  $\frac{(x-1)^2}{2}$ * (6.6) +  $\frac{(x-1)^3}{3*2}$ * 4 +  $\frac{(x-1)^4}{4*3*2}$ * 0 + . . . .

   y(x)  =  2.3 + (x-1) * (6.29) +  $\frac{(x-1)^2}{2}$ * (6.6) +  $\frac{(x-1)^3}{3*2}$ * 4 + 0

   y(x)  =  2.3 + (x-1) * (6.29) +  $\frac{(x-1)^2}{2}$ * (6.6) +  $\frac{(x-1)^3}{3*2}$ * 4

To find:

1. y at x = 1.1

=>  y(1.1)  =  2.3 + (1.1-1) * (6.29) +  $\frac{(1.1-1)^2}{2}$ * (6.6) +  $\frac{(1.1-1)^3}{3*2}$ * 4

               =  2.3 + (0.1) * (6.29) +  $\frac{(0.1)^2}{2}$ * (6.6) +  $\frac{(0.1)^3}{3*2}$ * 4

               = 2.3 + 0.629 + 0.33 + 0.00066

               = 3.2596

               ≈ 3.26

2. y at x = 1.2

=>  y(1.2)  =  2.3 + (1.2-1) * (6.29) +  $\frac{(1.2-1)^2}{2}$ * (6.6) +  $\frac{(1.2-1)^3}{3*2}$ * 4

               =  2.3 + (0.2) * (6.29) +  $\frac{(0.2)^2}{2}$ * (6.6) +  $\frac{(0.2)^3}{3*2}$ * 4

               = 2.3 + 1.258 + 0.132 + 0.00533

               = 3.69533

               ≈ 3.69

Therefore, y at 1.1 is approximately 3.26

and y at 1.2 is approximately 3.69