1. Mar 11, 2014

Marcoreus

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.

Last edited: Mar 11, 2014
2. Mar 11, 2014

Staff: Mentor

Your last equation is wrong. The equation before it is
ln y = λx + C
Exponentiating (making each side the exponent on e) gives
$$e^{ln y} = e^{λx + C} = e^{λx} \cdot e^C$$
Can you finish this?

What you have done is solve the DE using separation. I don't know what the "implicit Euler method" is, so you might not have solved this in the intended way.

3. Mar 11, 2014

Staff: Mentor

BTW, your title made me think you had a question about classical poetry. Odes are poems in praise of someone. ODEs are ordinary differential equations.

4. Mar 11, 2014

Marcoreus

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.

5. Mar 11, 2014

Simon Bridge

Welcome to PF;

Ode to ODEs

O My sweet ODE
How is it you can be
So beautiful and yet so deadly
To me

The method's Implicit Euler,
But deep-fried brain is oilier
Numerical iterations don't clean me up!
I'm soilier.

Variables separate
As we cogitate
But it's not the method no
But it's not the method no

By brain hurts so sweetly when you are around
Even when you make by head to pound
But O the joy when the general solution
is found.

Sweet ODE.

"implicit eulers method" is a numerical method involving iteration.
... oh you found it - good :)

6. Mar 11, 2014

Marcoreus

that was nice poem. i like it thanks

7. Mar 11, 2014

Simon Bridge

Imagine my disappointment when I found the thread was not about Odes :(
... these are all good questions.
Before we pitch in ere, how about having a go at answering them?