# Eulers Method to Approximate Differential Eq.

• brunie
In summary, to estimate y(1) using Euler's Method with 5 and 10 steps, we use the equation y_n+1 = y_n + h f(t_n , y_n) with a step size of 0.2 for 5 steps and 0.1 for 10 steps. The differential equation given is y'=2xy, and the answer to the DE is y(x) = e^(x^2). To get the values for f(t_n , y_n), we can use the function f(x,y)= 2xy or f(t,y)= 2ty. By plugging in the appropriate values, we can estimate y(1) to be close to e or 2.718

#### brunie

(** underscores refer to subscripts)

Consider inital value problem y'=2xy, y(0) = 1
Use Eulers Methos to estimate y(1) with 5 and 10 steps.

for 5 steps, increments are by 0.2
for 10 steps, increments are by 0.1

y_n+1 = y_n + h f(t_n , y_n)

where h is step size

the answer to the DE itself is y(x) = e^(x^2)
so these estimates should be close to e or 2.7182818...

i understand how to go from steps to step but i do not understand how to get the values for f(t_n , y_n)
all the examples I've looked at only had one variable so it is hard to compare

so for

y_1 = 1 + 0.2 f(t_1 , y_1)
y_2 = y_1 + 0.2 f(t_2 , y_2)
.
.

any help is appreciated

brunie said:
(** underscores refer to subscripts)

Consider inital value problem y'=2xy, y(0) = 1
Use Eulers Methos to estimate y(1) with 5 and 10 steps.

for 5 steps, increments are by 0.2
for 10 steps, increments are by 0.1

y_n+1 = y_n + h f(t_n , y_n)

where h is step size

the answer to the DE itself is y(x) = e^(x^2)
so these estimates should be close to e or 2.7182818...

i understand how to go from steps to step but i do not understand how to get the values for f(t_n , y_n)

One thing that may be causing trouble is that you have your algorithm in terms of "t" and "y" but your differential equation was given in terms of "x" and "y"! Either rewrite the equation as y'= 2ty or rewrite your algorithm in terms of x and y.
In any case, you are told that f(x,y)= 2xy or that f(t,y)= 2ty.

so for

y_1 = 1 + 0.2 f(t_1 , y_1)
= 1+ 0.2 (2(0)(1))= 1

y_2 = y_1 + 0.2 f(t_2 , y_2)
= 1+ 0.2(2(.2)(1))= 1.08

y_3= 1.08+ 0.2(2(.4)(1.08)= 1.08+ .1728= 1.2528

y_4= 1.2528+ 0.2(2(.6)(1.2528))= 1.5535 ...

If you are using MATLAB I have some code which can get you started.