Fint the modified equation when the implicit Euler method is applied to y'= f(y). If f(y)=λy, where λ is negative. what is the effect on the amplication factor?

=>

y ' = λ * y

dy / dx = λ * y

dy / y = λ dx

ln y = λ* x + C

y = Ae^( λ* x ), the constant factor does not depend on λ.

i SOLVE THIS FOR THE ACTUAL SOLUTIONS

NOW,

The implicit Euler scheme is given by:

y_(n+1)= y_n +hf(y_n+1 , t_n+1 )

For f(y)=λ y, we have:

y_(n+1)= y_n +hf(y_n+1 , t_n+1 )= y_n + h λ y_n+1

Solving this for y_n+1 (in general, this is not possible), we arrive at:

y_n+1 = y_n / (1-h λ)..............(eqn 1)

From (eqn 1), we can see that if |1−h λ|≥1, the solution is decaying (stable). Compare this to the actual solution of y(x)=Ae^(λ x).

If we have λ being negative, we would have:

y_n+1 = y_n / (1+h λ)..............(eqn 2)

Compare this to the actual solution of y(x)=Ae^(−λ x). What conclusion can i draw?

Trying it for λ= ±1 , what happens to stability.