For what $h$ is Euler's explicit method stable?

For which value of the discretization step $$h$$ the Euler explicit method applied to the differential equation $$y'(x)+\frac{1}{4}y(x)=x$$ with the initial condition $$y(1)=1$$ is stable?

I wrote the equation as $$y'(x)=-\frac{1}{4}y(x)+x$$ and put the function into equation of explicit Euler's method: $$y_{n+1} = y_{n} + h\left(x_{n}-\frac{1}{4}y_n\right)$$ But I'm stuck and don't know what more to do. I know that something has to be $$<1$$ for it to be a stable method, but don't know which part of the equation.

Possible answers: $$\frac{1}{4}, \frac{2}{3}, \frac{1}{2}, 1$$.

Solving the recurrence

$$y_{n+1}-\left(1-\frac h4\right)y_n = h x_n$$

we got easily

$$y_n = \left(1-\frac h4\right)^{n-1}\left(c_1 +h\sum_{k=0}^{n-1}\left(1-\frac h4\right)^{-k}x_k\right)$$

hence the condition is

$$|1-\frac h4| \lt 1$$

NOTE

Here $$x_n = 1 + nh$$ so with the condition $$y_0 = 1, x_0 = 1$$ we have $$c_1 = 1-\frac h4$$ so we have finally

$$y_n = 13 \left(1-\frac{h}{4}\right)^n+4 h n-12$$

and as we can verify, $$y_n$$ approaches $$y(x)$$ maintaining stability for $$0 < h < 8$$

• All of the possible answers fit into condition and that's my main problem :( I need to write exact value of h. Jan 21 at 19:55