Question

Use Euler's method with step size 0.1 to estimate y(2.5), where y(x) is the solution of the initial-value problem y' = 3y + 2xy, y(2) = 1. (Round your answer to four decimal places.) y(2.5) =

Answer 1

Answer:

1.5939 $($Correct upto four desimal places$)$

Step-by-step explanation:

To obtain the large value at $x_n=2.5$ with small step length $h=0.1$ and $x_0=2,y_0=1,x_n=2.5$ , where $x_n=x_0+nh$ we have to use c-programme.  The c-programme for Eular method is given below,

$#$include$<$stdio.h$>$

float fun$($float$x$, float$y)$

${$

float $f;$

$f=3\times y+2\times x\timesy;$

return $f;$

$}$

main$()$

${$

float $a,b,x,y,h,t,k;$

printf$("\nEnter$ $x_0,y_0,h,xn:");$

scanf$("%f%f%f%f",&a,&b,&h,&t);$

$x=a;$

$y=b;$

printf$("\n x \t y \t");$

while$(x<=t)$

${$

$k=h\times fun(x,y);$

$y=y+k;$

$x=x+h;$

printf$("%0.3f\t%0.3f\n",x,y);$

$}$

$}$

which will give required answer $y(2.5)=$, correct upto four desimal places.

Answer 2

The estimated solution of the initial-value problem is $y(2.5) = 6.7809$.

How to use Euler's method

The Euler's method is a multistage numerical method, to estimate a point of a given solution based on a given initial value. The expressions required in this method are presented below:

$f(x_{i}, y_{i}) = \frac{dy}{dx}|_{P_{i}}$     (1)

$x_{i+1} = x_{i}+h$     (2)

$y_{i+1} = y_{i}+h\cdot f(x_{i}, y_{i})$ (3)

Where $h$ is the step size.

Now we start iterating this differential equation:

Iteration 1

By (1):

$f(x_{o},y_{o}) = 3\cdot (1) + 2\cdot (2)\cdot (1)$

$f(x_{o},y_{o}) = 7$

By (2):

$x_{1} = 2 + 0.1$

$x_{1} = 2.1$

By (3):

$y_{1} = 1 + 0.1\cdot 7$

$y_{1} = 1.7$

Iteration 2

By (1):

$f(x_{1},y_{1}) = 3\cdot (1.7) + 2\cdot (2.1)\cdot (1.7)$

$f(x_{1},y_{1}) = 12.24$

By (2):

$x_{2} = 2.1 + 0.1$

$x_{2} = 2.2$

By (3):

$y_{2} = 1 .7+ 0.1\cdot 12.24$

$y_{2} = 2.924$

Iteration 3

By (1):

$f(x_{2},y_{2}) = 3\cdot (2.924) + 2\cdot (2.2)\cdot (2.924)$

$f(x_{2},y_{2}) = 21.638$

By (2):

$x_{3} = 2.2 + 0.1$

$x_{3} = 2.3$

By (3):

$y_{3} = 2.2+ 0.1\cdot 21.638$

$y_{3} = 4.3638$

Iteration 4

By (1):

$f(x_{3},y_{3}) = 3\cdot (4.3638) + 2\cdot (2.3)\cdot (4.3638)$

$f(x_{3},y_{3}) = 33.165$

By (2):

$x_{4} = 2.3 + 0.1$

$x_{4} = 2.4$

By (3):

$y_{4} = 2.3+ 0.1\cdot 33.165$

$y_{4} = 5.6165$

Iteration 5

By (1):

$f(x_{4},y_{4}) = 3\cdot (5.6165) + 2\cdot (2.4)\cdot (5.6165)$

$f(x_{4},y_{4}) = 43.8087$

By (2):

$x_{5} = 2.4 + 0.1$

$x_{5} = 2.5$

By (3):

$y_{5} = 2.4+ 0.1\cdot 43.8087$

$y_{5} = 6.7809$

The estimated solution of the initial-value problem is $y(2.5) = 6.7809$. $\blacksquare$

To learn more on Euler's method, we kindly invite to check this verified question: https://brainly.com/question/11325462