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.