# Euler backward method

The ODE that leads to Figure shown is $y' = 10y$. The exact solution satisfies $y(0) = 1$, and Euler’s method is applied with step size $h = 0.1$. What are the initial values $y(0)$ for the other two trajectories, depicted in Figure in dashed lines? My attempt - I know how to do the forward Euler method $$y(h) = y_1 = y_0 + h*f(x_0, y_0)$$

So, we have $y_1 = 1 + 0.1(10*1) = 2$

Similarly, $y_2 = 2 + 0.1(10*2) = 4$

However, I do not know how to backtrack them to get $y(0)$ Any help is appreciated!

• Are you sure you're supposed to use a numerical method to figure this out? The backward Euler method isn't really meant to help you move "backward" in time towards $t=0$. I suppose you could do that with any similar numerical method if $h < 0$, but it still just seems strange. – tilper Apr 27 '17 at 19:40

## 1 Answer

Consider the first of the two dashed solution (the one just above the red curve). It is expressed as follows:

$$\bar{y}(x) = \bar{y}_0 e^{10x}.$$

Since you know that for $x=0.1$, the value of this function is $2$, then:

$$2 = \bar{y}_0 e^{10 \cdot 0.1} \Rightarrow \bar{y}_0 = \frac{2}{e}.$$

For the second curve, since for $x=0.2$ it passes from $4$, then:

$$4 = \bar{y}_0 e^{10 \cdot 0.2} \Rightarrow \bar{y}_0 = \frac{4}{e^2}.$$

• Yes, I think this is the intent. – tilper Apr 27 '17 at 19:42