Dif EQ

**caeder012** · November 10th 2007, 12:55 PM

So I'm trying to solve this partially decoupled system:

$\frac{dx}{dt} = 2x + 3y$

$\frac{dy}{dt} = -4y$

1.) Solve it WITHOUT using matrices. (HINT: start with the second equation)

2.) Solve it with using matrices.

I can do #2... but #1 is so hard.

For #2, I set up the system into a matrix so I get X' = AX, with A being:

[[2,3],[0,-4]]

Then, I found the eigenvalues:

$\lambda_1 = 2, \lambda_2 = -4$

And the corresponding eigenvectors (call them $k_1, k_2$ , respectively:

k_1 = [[1],[0]], k_2 = [[1/2],[1]

So then the solution is:

$X = c_1e^{2t}k_1 + c_2e^{-4t}k_2$

Voila.

Now for #1...

So solving the 2nd equation first...:

$\frac{dy}{-4y} = dt \implies \frac{-ln(|y|)}{4} = t + C$

Solve for y (I ignored the absolute value so I don't get 2 diff soln's):

$y = e^{-4t-4c}$

So now plug this into 1st eq.:

$\frac{dx}{dt} = 2x + 3\cdot e^{-4t-4c}$

No idea how to solve this! Maybe I did something wrong.. no idea.

**topsquark** · November 10th 2007, 01:14 PM

Originally Posted by caeder012

:

$\frac{dx}{dt} = 2x + 3y$

$\frac{dy}{dt} = -4y$

So solving the 2nd equation first...:

$\frac{dy}{-4y} = dt \implies \frac{-ln(|y|)}{4} = t + C$

Solve for y (I ignored the absolute value so I don't get 2 diff soln's):

$y = e^{-4t-4c}$

So now plug this into 1st eq.:

$\frac{dx}{dt} = 2x + 3\cdot e^{-4t-4c}$

No idea how to solve this! Maybe I did something wrong.. no idea.

You can ignore the absolute value bars only so long as y is an always positive function. Since it is (barring an arbitrary constant making it negative) don't worry about it.

$\frac{dx}{dt} = 2x + 3Ae^{-4t}$ <-- Defining $A = e^{-4c}$ for convenience

$\frac{dx}{dt} - 2x = 3Ae^{-4t}$

Solve the homogeneous equation first:
$\frac{dx_h}{dt} - 2x_h = 0$

The characteristic polynomial for this is $m - 2 = 0$ , so
$x_h = Be^{2t}$

For the particular equation
$\frac{dx_p}{dt} - 2x_p = 3Ae^{-4t}$
we can try a solution
$x_p = De^{-4t}$

Using this gives:
$-4De^{-4t} - 2De^{-4t} = 3Ae^{-4t}$

$-6D = 3A$

Thus
$x_p = -\frac{A}{2} e^{-4t}$

So the overall solution is
$x = x_h + x_p = Be^{2t} - \frac{A}{2} e^{-4t}$

which matches your other solution.

-Dan

**CaptainBlack** · November 10th 2007, 01:16 PM

Originally Posted by caeder012

So now plug this into 1st eq.:

$\frac{dx}{dt} = 2x + 3\cdot e^{-4t-4c}$

No idea how to solve this! Maybe I did something wrong.. no idea.

Lets assume your work up to this point is correct. Rewrite what you have as:

$\frac{dx}{dt} -2x = A\cdot e^{-4t}$

This is an inhomogeneous linear constant coefficient ODE, and its general
solution is the sum of the general solution of the homogeneous equation:

$\frac{dx}{dt} -2x = 0$

and any particular integral of the original equation.

I will assume you know how to solve the homogeneous equation, and that its
solution is:

$x(t)=C e^{2t}$

Now to find a particular solution of the inhomegeneous equation you try a trial
solution of the form $x(t)=B~e^{-4t}$ .

RonL

**Soroban** · November 10th 2007, 01:42 PM

Hello, caeder012!

You're making the problem very messy . . .

$\frac{dx}{dt} = 2x + 3y$ .[1]
$\frac{dy}{dt} = -4y$ .[2]

1) Solve it WITHOUT using matrices. (HINT: start with the second equation)

From [2], we have: . $\frac{dy}{y} \:=\:-4\,dt$

. . Integrate: . $\ln y \:=\:-4t + c$

. . Then: . $y \:=\:e^{-4t+c} \:=\:e^{-4t}\cdot e^c \quad\Rightarrow\quad\boxed{ y\:=\:C_1e^{-4t}}$

Substitute into [1]: . $\frac{dx}{dt} \:=\:2x + 3\left(C_1e^{-4t}\right)$

. . Then we have: . $\frac{dx}{dt} - 2x \:=\:3C_1e^{-4t}$

Integrating factor: . $I \;=\;e^{\int(-2)\,dt} \;=\;e^{-2t}$

Multiply by $I\!:\;\;e^{-2t}\!\cdot\!\frac{dx}{dt} - 2xe^{-2t} \;=\;3C_1e^{-6t}$

. . and we have: . $\frac{d}{dt}\left(e^{-2t}\!\cdot\!x\right) \;=\;3C_1e^{-6t}$

Integrate: . $e^{-2t}\!\cdot\!x \;=\;3C_1\int e^{-6t}dt \quad\Rightarrow\quad e^{-2t}x \;=\;-\frac{1}{2}C_1e^{-6t} + C_2$

Multiply by $e^{2t}\!:\;\;\boxed{x \;=\;-\frac{1}{2}C_1e^{-4t} + C_2e^{2t}}$

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