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Advanced Topic in Odes 1

grandy

Member
Dec 26, 2012
73
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 ' = lambda * y
dy / dx = lambda * y
dy / y = lambda dx
ln y = lambda * x + C
y = [ C / lambda ]e^( lambda * x )

This is what I have done so far. Can anyone help me if Iam going right direction to answer the question?


 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
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 ' = lambda * y
dy / dx = lambda * y
dy / y = lambda dx
ln y = lambda * x + C
y = [ C / lambda ]e^( lambda * x )

This is what I have done so far. Can anyone help me if Iam going right direction to answer the question?


I agree with the solution up to $\displaystyle \begin{align*} \ln{|y|} = \lambda x + C \end{align*}$, from there you should get $\displaystyle \begin{align*} y = A e^{\lambda x } \end{align*}$, the constant factor does not depend on $\displaystyle \begin{align*} \lambda \end{align*}$...
 

grandy

Member
Dec 26, 2012
73
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.