Solve numerical differential equation using Modified Euler method (1st order derivative) calculator

SolutionHelpInput functions

Solution

Solution provided by AtoZmath.com

1. Find y(0.5) for `y'=-2x-y`, y(0) = -1, with step length 0.1

2. Find y(2) for `y'=(x-y)/2`, y(0) = 1, with step length 0.2

3. Find y(0.3) for `y'=-(x*y^2+y)`, y(0) = 1, with step length 0.1

4. Find y(0.2) for `y'=-y`, y(0) = 1, with step length 0.1

Example

1. Find y(0.2) for `y'=(x-y)/2`, `x_0=0, y_0=1`, with step length 0.1 using Modified Euler method (1st order derivative)

Solution:
Given `y'=(x-y)/(2), y(0)=1, h=0.1, y(0.2)=?`

Here, `x_0=0,y_0=1,h=0.1,x_n=0.2`

`y'=(x-y)/(2)`

`:. f(x,y)=(x-y)/(2)`

Modified Euler method
`y_(m+1)=y_m+hf(x_m+1/2 h,y_m + 1/2 hf(x_m,y_m))`

`f(x_0,y_0)=f(0,1)=-0.5`

`x_0+1/2 h=0+0.1/2 =0.05`

`y_0 + 1/2 hf(x_0,y_0)=1+0.1/2 * -0.5=0.975`

`f(x_0+1/2 h, y_0 + 1/2 hf(x_0,y_0)=f(0.05,0.975)=-0.4625`

`y_1=y_0+hf(x_0+1/2 h, y_0 + 1/2 hf(x_0,y_0))=1+0.1*-0.4625=0.9538`

`:.y(0.1)=0.9538`

Again taking `(x_1,y_1)` in place of `(x_0,y_0)` and repeat the process

`f(x_1,y_1)=f(0.1,0.9538)=-0.4269`

`x_1+1/2 h=0.1+0.1/2 =0.15`

`y_1 + 1/2 hf(x_1,y_1)=0.9538+0.1/2 * -0.4269=0.9324`

`f(x_1+1/2 h, y_1 + 1/2 hf(x_1,y_1)=f(0.15,0.9324)=-0.3912`

`y_2=y_1+hf(x_1+1/2 h, y_1 + 1/2 hf(x_1,y_1))=0.9538+0.1*-0.3912=0.9146`

`:.y(0.2)=0.9146`

`n`	`x_n`	`y_n`	`x_(n+1)`	`y_(n+1)`
0	0	1	0.1	0.9538
1	0.1	0.9538	0.2	0.9146

Input functions

Sr No.	Function	Input value
1.	`x^3`	x^3
2.	`sqrt(x)`	sqrt(x)
3.	`root(3)(x)`	root(3,x)
4.	sin(x)	sin(x)
5.	cos(x)	cos(x)
6.	tan(x)	tan(x)
7.	sec(x)	sec(x)
8.	cosec(x)	csc(x)
9.	cot(x)	cot(x)
10.	`sin^(-1)(x)`	asin(x)
11.	`cos^(-1)(x)`	acos(x)
12.	`tan^(-1)(x)`	atan(x)
13.	`sin^2(x)`	sin^2(x)
14.	`log_y(x)`	log(y,x)
15.	`log_10(x)`	log(x)
16.	`log_e(x)`	ln(x)
17.	`e^x`	exp(x) or e^x
18.	`e^(2x)`	exp(2x) or e^(2x)
19.	`oo`	inf