Differential equations with Euler's method A differential equation y' + 2y = 2 - e^(-4*t)
With starting point y(0) = 1 and increment 0,1.
I have to find the approximate value of the function in time t = {0,1; 0,2; 0,3; 0,4; 0,5; }
And I have to compare the values with the correct answer: 
y(t) = 1 + 0,5 * e^(-4*t) - 0,5 * e^(-2*t)

There is also this equation given, I think it's not a part of the question, but I'll add it:
y' = f(t,y) = 2 - e^(-4*t) - 2 * y

The solution is given like this:
Time - t | Approximation | Correct answer | Error [%]
0        | 1             | 1              | 0
0,1      | 0,9           | 0,9257946      | 2,79
0,2      | 0,852968      | 0,8895045      | 4,11
0,3      | 0,8374415     | 0,8761913      | 4,42
0,4      | 0,8398338     | 0,8762834      | 4,16
0,5      | 0,8516774     | 0,8837292      | 3,63

I've tried looking up online about Eurler's method, but I still can't understand where all these numbers come from. There is only one formula given which is x(t+h) = x(t) + h*x'(t) by the professor. The subject is called Numerical Methods.
 A: Use 
y(t+h) = y(t) + h*f(t,y(t)),

you have the parts all in the question.

The explicit form $y'=f(t,y)$ is the standard form that is at the base of most numerical solution methods, you get $y'$ isolated from the original equation by subtracting $2y$ from both sides of the equation.
A: I assume you know that y' is a standard notation for the derivative, also written as $\frac{dy}{dt}$ and defined as $\lim_{\Delta t\to 0}\frac{\Delta y}{\Delta t}$ where "$\Delta y$" is the change in y that results from the small change in x, $\Delta t$.  The idea behind "Euler's method" is to approximate the derivative by $\frac{\Delta f}{\Delta t}$.  That is, the equation, $y'+ 2y= 2- e^{-4t}$ can be approximate by $\frac{\Delta y}{\Delta t}= -2y+ 2- e^{-4t}$. 
Write that as $\Delta y= (-2y+ 2- e^{-4t})\Delta x$.  Since you are told to use "t= 0.1; 0.2; 0.3; 0.4; 0.4" $\Delta t= 0.1$ since t increases by 0.1 at each step. Further, you are told that y(0)= 1.  With t= 0, y= 1, $-2y+ 2- e^{-4t} is $-2(1)+ 2- e^0= -1.  Since $\Delta x= 0.1$, $\Delta y= (-2y+ 2- e^{-4t})\Delta t= (-1)(0.1)= -0.1$.  That is, while x increases by 0.1, from 0 to 0.1, y "increases" by -0.1 or decreases for 1 to 1- 0.1= 0.9.  That's where the "0.1" and "0.9" (your "0,1" and "0,9") come from in the second line.  Of course, the "0,9257946" is calculated putting t= 0.1 into the given "correct solution" and the % error is the difference between 0,9257946 and 0,9= 0,0257946, divided by the "correct" 0,9257946.
Now repeat. 
