# Math Help - Coupled differential equations

1. ## Coupled differential equations

I really need help solving these coupled differential equations

$\frac{dZ}{dt} = -aZ$

$\frac{dY}{dt} = cZ - gY$

where a = b+c+g and a,b,c,g are all constants.

The initial conditions are

$Z(0) = Z_0$, and $Y(0) = Y_0$,
I know that $Z(t) = e^{-at}$ , but im having trouble with finding Y(t)

This is a past exam question so I have the answer already, but I need to know exactly how to do it for a homework question I have.
Here is the solution

$Y(t) = Y_0 e^{-gt} + \frac{c*Z_0*(e^{-gt} - e^{-at}) }{a - g}$

2. I only remember a few steps in these kinds of problems...
BTW Z = Z_0 e^{-at} not just e^{-at}
Y = PI + CF = particular integral + complementary function I believe
You're given
Dy = cZ - gY where D = differentiation operator so
(D+g) Y = cZ
PI (or CF) is given by (D+g) Y = 0 i.e. PI = const * e^{-gt}
CF (or PI) is given by (D+g) Y = cZ = he^{-at} where h = cZ_0
Assume CF= ve^{-wt} then (D+g) Y = v(g-w)e^{-wt} , v w const
Eqauating powers of t, w = a so CF = ve^{-at}
v(g-a)e^{-at} = cZ_0e^{-at} so v = cZ_0/(g-a)
Now Y = PI + CF = ke^{-gt} + cZ_0/(g-a) e^{-at}
Almost there, hopefully someone more knowledgeable will correct my missteps here

3. Originally Posted by Maccaman
I really need help solving these coupled differential equations

$\frac{dZ}{dt} = -aZ$

$\frac{dY}{dt} = cZ - gY$

where a = b+c+g and a,b,c,g are all constants.

The initial conditions are

$Z(0) = Z_0$, and $Y(0) = Y_0$,
I know that $Z(t) = e^{-at}$ , but im having trouble with finding Y(t)

This is a past exam question so I have the answer already, but I need to know exactly how to do it for a homework question I have.
Here is the solution

$Y(t) = Y_0 e^{-gt} + \frac{c*Z_0*(e^{-gt} - e^{-at}) }{a - g}$
You're very close for the solution to $Z(t)$. $Z(t)={\color{red}C}e^{-at}$

Since $Z(0)=Z_0,~Z_0=C\implies C=Z_0$

So $Z(t)=Z_0e^{-at}$

Now plug your solution for Z into the differential equation you have for Y:

$\frac{\,dY}{\,dt}=cZ_0e^{-at}-gY$

Rewrite the equation to see that its a linear equation:

$\frac{\,dY}{\,dt}+gY=cZ_0e^{-at}$

The integrating factor is $\varrho(t)=e^{\int g\,dt}=e^{gt}$

Thus, we see that $\frac{d}{\,dt}\left[Ye^{gt}\right]=cZ_0e^{(g-a)t}$

Integrating both sides yields:

$Ye^{gt}=\frac{cZ_0}{g-a}e^{(g-a)t}+C\implies Y=\frac{cZ_0}{g-a}e^{-at}+Ce^{-gt}$

Applying the initial condition $Y(0)=Y_0$, we see that $Y_0=\frac{cZ_0}{g-a}+C\implies C=Y_0-\frac{cZ_0}{g-a}$

Thus, $Y=\frac{cZ_0}{g-a}e^{-at}+\left[Y_0-\frac{cZ_0}{g-a}\right]e^{-gt}$

Therefore,

$\color{red}\boxed{Y(t)=Y_0e^{-gt}+\frac{cZ_0\left[e^{-gt}-e^{-at}\right]}{a-g}}$

Does this make sense?

--Chris

4. Originally Posted by Chris L T521
You're very close for the solution to $Z(t)$. $Z(t)={\color{red}C}e^{-at}$

Since $Z(0)=Z_0,~Z_0=C\implies C=Z_0$

So $Z(t)=Z_0e^{-at}$
What I originally wrote was a typo. I did mean to type $Z(t)=Z_0e^{-at}$

But thanks for providing me with the details for obtaining Y(t). It makes perfect sense. You

5. ## another way

or the system:

$
\frac{dx}{dt}=5x+3y$

$\frac{dy}{dt}=x+7y

$

You determined the eigenvalues and eigenvectors. Thus the solution is:

$
\left(
\begin{array}{c} x(t) \\y(t)\end{array}\right)
=k_1e^{4t}
\left(\begin{array}{c} 3 \\ -1 \end{array}\right)+k_2e^{8t}
\left(\begin{array}{c} 1 \\ 1 \end{array}
\right)
$

Right?

Substituting the initial conditions:

$

\left(\begin{array}{c} 5 \\ 1 \end{array}\right)=
k_1\left(\begin{array}{c}3 \\ -1 \end{array}\right)+
k_2\left(\begin{array}{c}1 \\ 1 \end{array}\right)
$

Now just figure out what the constants are. You know how to read this matrix stuf huh? That last one is just:

$
5=3k_1+k2$

$1=-k_1+k_2
$

Same dif for reading the other one.

Edit: Oh yea:

Welcome to PF and

"Differential Equations" by Blanchard, Devaney, and Hall is one DE book I use.

6. Originally Posted by ajbriggs
or the system:

$

\frac{dx}{dt}=5x+3y

\frac{dy}{dt}=x+7y

$

You determined the eigenvalues and eigenvectors. Thus the solution is:

$
\left(\begin{array}{c} x(t) \\y(t)\end{array}\right)=k_1e^{4t}
\left(\begin{array}{c} 3 \\ -1 \end{array}\right)+k_2e^{8t}
\left(\begin{array}{c} 1 \\ 1 \end{array}\right)
$

Right?

Substituting the initial conditions:

$

\left(\begin{array}{c} 5 \\ 1 \end{array}\right)=
k_1\left(\begin{array}{c}3 \\ -1 \end{array}\right)+
k_2\left(\begin{array}{c}1 \\ 1 \end{array}\right)
$

Now just figure out what the constants are. You know how to read this matrix stuf huh? That last one is just:
$
5=3k_1+k2

1=-k_1+k_2
$

Same dif for reading the other one.

Edit: Oh yea:

Welcome to PF and

"Differential Equations" by Blanchard, Devaney, and Hall is one DE book I use.
PF?