The modified Euler's method is $w_{0}=\alpha$, $$w_{n+1}=w_{n}+\frac{h(f(w_{n},t_{n})+f(w_{n}+hf(w_{n},t_{n}),t_{n+1}))}{2}$$.
Apply this method to the IVP $$y^\prime=\lambda y,$$ $y(0)=1$, with $\lambda<0$, and find the conditions on $\lambda$ and $h$ which ensures $w_{n}\rightarrow 0$ as $n\rightarrow \infty$.
From the comments:
I have tried to find a general expression for $w_n$ and get the following $w_n=((h\lambda)^2+2h\lambda+2)^{n+\frac{1}{2^n}}$. But don't know how to move forward from here
$y_{n+1}=(\frac{\lambda^2 h^2+2\lambda h+2}{2})^{n+1}y_{0}$. Now given the fact that $\lambda<1$, the growth factor ($y_{n+1}/y_{0}$), has to be less than $1$, in order for $w_{n}\rightarrow 0$, as $n\rightarrow \infty$. Now, $\frac{\lambda^2 h^2+2\lambda h+2}{2}<1,\implies (\lambda h)^2+2\lambda h<0,\implies \lambda h(\lambda h+2)<0,\implies -2<\lambda h<0, \implies h<\frac{-2}{\lambda},\lambda<0$