Euler's method for first three approximations? 
I have tried variations of the problem for an hour at least and cannot get around to sloving this one.
Thank you for input!
 A: You can derive Euler's method from Taylor series as

$$ y(x_0+h)\approx y(x_0)+y'(x_0)h,\quad h=\Delta x.  $$

Put $x_1=x_0+h$ in the above equation which gives

$$ y(x_1)\approx y(x_0)+y'(x_0)h,\quad h=\Delta x.  $$

If we repeat the process we get

$$ y(x_2)=y(x_1+h)\approx y(x_1)+y'(x_1)h,\quad h=\Delta x , $$

which can be generalized to 

$$ y(x_n)=y(x_{n-1}+h)\approx y(x_{n-1})+y'(x_{n-1})h,\quad h=\Delta x \longrightarrow (*). $$

In your case, we will find $y(x_1)$ and you will do the rest. We have given the following information

$$ x_0=0,\quad y(0)=5,\quad h=\Delta x = .1,\quad y'(x)=-7x^6e^{-x^7}\implies y'(0)=0. $$

Now substitute in
$$ y(x_1)=y(x_0+h)\approx y(x_0)+y'(x_0)h,\quad h=\Delta x  $$
gives
$$ y(x_1)=y(0+0.1)\approx y(0)+y'(0)(0.1),\quad h=\Delta x $$

$$ y(x_1)=y(0.1) \approx 5 + 0 =5 \implies y(x_1)\approx 5. $$

To find $y(x_2)$ you need the following by checking the formula $(*)$

$$ h=0.1,\quad x_1 = x_0 + h = 0.1,\quad y(x_1)\approx 5,\quad y'(x_1)=y'(0.1)=-7(0.1)^6e^{-(0.1)^7}.  $$

I think you can continue now.
A: Euler's method is an iterative method using the linear approximation $y_{n+1}=y_n+h\cdot y'(x)$ where $h$ is the step size (given as $dx$ in the problem).
We have $y_0=5$, can you use the approximation to get $y_1$?
As for the exact case, that differential equation is separable.
