# Thread: differencial System

1. ## differencial System

Suppose you are given the following system of linear dif eq's:

$\frac{dx}{dt} = -\lambda_1 x$
$\frac{dy}{dt} = \lambda_1 x - \lambda_2 y$
$\frac{dz}{dt} = \lambda_2 y$

Find a solution for this system WITHOUT USING matrices or the method using differential operators for elimination.

2. Originally Posted by alikation0
Suppose you are given the following system of linear dif eq's:

$\frac{dx}{dt} = -\lambda_1 x$
$\frac{dy}{dt} = \lambda_1 x - \lambda_2 y$
$\frac{dz}{dt} = \lambda_2 y$

Find a solution for this system WITHOUT USING matrices or the method using differential operators for elimination.
Solve the first, for $x(t)$, then use that solution to solve the second for $y(t)$, then use both solutions to solve the third for $z(t)$. You will have to handle the cases $\lambda_1=\lambda_2$ and $\lambda_1 \ne \lambda_2$ separately.

RonL

3. Originally Posted by CaptainBlack
Solve the first, for $x(t)$, then use that solution to solve the second for $y(t)$, then use both solutions to solve the third for $z(t)$. You will have to handle the cases $\lambda_1=\lambda_2$ and $\lambda_1 \ne \lambda_2$ separately.

RonL
Ok so I'm assuming it's ok the represent the first as:

$\frac{dx}{dt} = -\lambda_1 x(t)$

How would I solve this for x(t):

$dx = -\lambda_1 x(t)dt$
$\frac{1}{x(t)}dx = -\lambda_1 dt$

$ln(x(t)) = -\lambda dt$

This looks like an extremely hard problem having to deal with if $\lambda_1 = \lambda_2$ etc...

4. Okay:

Apparently your way is correct Captain! I was told that I would want to treat the lambda's separately. However, the TA said it was fine if I only do case where I assumed the lambda's were NOT equal and proceeded there (so I don't have to do the case where they).

Could you help me? I got to:

$ln(x(t)) = \lambda_1 dt$
$ln(x(t)) = \lambda_1 t$

$ln(x(t))/t = \lambda_1$

So then according to what you said:

$\frac{dy}{dt} = \frac{ln(x(t))}{t} x(t) - \lambda_2 y(t)$

5. Originally Posted by alikation0
Okay:

Apparently your way is correct Captain! I was told that I would want to treat the lambda's separately. However, the TA said it was fine if I only do case where I assumed the lambda's were NOT equal and proceeded there (so I don't have to do the case where they).

Could you help me? I got to:

$ln(x(t)) = \lambda_1 dt$
$ln(x(t)) = \lambda_1 t$

$ln(x(t))/t = \lambda_1$

So then according to what you said:

$\frac{dy}{dt} = \frac{ln(x(t))}{t} x(t) - \lambda_2 y(t)$

You misunderstood what CaptainBlack was saying. You are looking for x(t) in the first step, not $\lambda _1$.

$\frac{dx}{dt} = \lambda _1 x$

$\frac{dx}{x} = \lambda _1 dt$

$ln(x) = \lambda _1 t + C$

$x(t) = Ae^{\lambda _1 t}$
where $A = e^C$.

Now plug this into the second equation:
$\frac{dy}{dt} = \lambda _1 x - \lambda _2 y$

$\frac{dy}{dt} = \lambda _1 Ae^{\lambda _1 t} - \lambda _2 y$

Now solve for y(t).

-Dan