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- #1

=>

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?

- Thread starter grandy
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- Thread starter
- #1

=>

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 $\displaystyle \begin{align*} \lambda \end{align*}$...

=>

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?

- Thread starter
- #3

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.