# First order linear differential equations and an application

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• Feb 9th 2008, 01:26 PM
dark_knight_307
First order linear differential equations and an application
If anyone can help me integrate the following first order differential equation, I'd appreciate it!...

dy=(y(tan x) + 2e^x)dx

I got it this far....

y = (C/abs(cos x))*(integral of e^x*abs(cos x)*dx)

And here's another problem....

An object falling near the earth's surface encounters air resistance that is proportional to its velocity. The acceleration due to gravity is -9.8m/s^2. So, without air resistance the object's acceleration can be modeled by the differential equation. dv/dt = -9.8. But aerodynamic drag represents a considerable retarding force as velocity increases. Thus a better model for an object falling near the surface of the earth is: dv/dt = kv - 9.8, where k is a constant of proportionality.

1. What are the units of k? Is k positive or negative?
2. Find the velocity of the object as a function of time if the initial velocity is V sub 0.
3. Use #2 to find the limit of the velocity as t -> infinity.
4. Find the position function s(t) of the object.
• Feb 9th 2008, 01:55 PM
Krizalid
Quote:

Originally Posted by dark_knight_307
dy=(y(tan x) + 2e^x)dx

Do you know what the integrating factor is?

Your equation is after make-up $y' - y\tan x = 2e^x .$ So our integrating factor will be

$\mu (x) = \exp \left\{ { - \int {\tan x\,dx} } \right\} = \cos x.$ Now we just multiply the entire equation by $\mu (x),$ this yields

$y'\cos x - y\sin x = 2e^x \cos x \implies (y\cos x)' = 2e^x \cos x.$

From there you may perform integration and you're done.

(See my signature for LaTeX typesetting.)
• Feb 9th 2008, 02:04 PM
mr fantastic
Quote:

Originally Posted by dark_knight_307
If anyone can help me integrate the following first order differential equation, I'd appreciate it!...

dy=(y(tan x) + 2e^x)dx

I got it this far....

y = (C/abs(cos x))*(integral of e^x*abs(cos x)*dx)

[snip]

You don't need the absolute value since $e^{\ln |\cos x|} =$ plain old $\cos x$.

The solution is actually $y = \frac{1}{\cos x} \int \cos x \, e^x \, dx$. There is no arbitrary constant out the front.

I personally like to put in the arbitrary constant in advance .......

$y = \frac{1}{\cos x} \left [\int \cos x \, e^x \, dx + C \right] = \frac{1}{\cos x} \int \cos x \, e^x \, dx + \frac{C}{\cos x}$.

To find the integral, an obvious (with the exception of Krizalid ;) ) aproach would be to use integration by parts.
• Feb 9th 2008, 02:05 PM
dark_knight_307
How do you integrate 2e^x cos x?
• Feb 9th 2008, 02:11 PM
Krizalid
mr fantastic already told ya that: integrate by parts. (Twice.)

Another attempt could be using complex numbers.
• Feb 9th 2008, 02:26 PM
dark_knight_307
Quote:

Originally Posted by Krizalid
mr fantastic already told ya that: integrate by parts. (Twice.)

What does that mean?
• Feb 9th 2008, 02:27 PM
Krizalid
Do you know what integration by parts technique is?

How is then you're covering ODE if you don't know how to integrate by parts?
• Feb 9th 2008, 02:28 PM
dark_knight_307
Quote:

Originally Posted by Krizalid
Do you know what integration by parts technique is?

How is then you're covering ODE if you don't know how to integrate by parts?

Could you show me how to do it, really fast like?
• Feb 9th 2008, 02:31 PM
mr fantastic
Quote:

Originally Posted by dark_knight_307
[snip]

And here's another problem....

An object falling near the earth's surface encounters air resistance that is proportional to its velocity. The acceleration due to gravity is -9.8m/s^2. So, without air resistance the object's acceleration can be modeled by the differential equation. dv/dt = -9.8. But aerodynamic drag represents a considerable retarding force as velocity increases. Thus a better model for an object falling near the surface of the earth is: dv/dt = kv - 9.8, where k is a constant of proportionality.

1. What are the units of k? Is k positive or negative? Mr F says: Units on left have to match units on right. So kv needs to be m/s^2. And you know v is m/s .... Downwards is -ve, upwards is +ve. Gravity is down and v is down. But kv is in oposite direction to gravity ......

2. Find the velocity of the object as a function of time if the initial velocity is V sub 0. Mr F says: Solve the DE: $\frac{dt}{dv} = \frac{1}{kv - 9.8}$ .......

3. Use #2 to find the limit of the velocity as t -> infinity. Mr F says: Take the limit in your answer. Check: Is it the same as you get by solving 0 = kv - 9.8 (got by putting dv/dt = 0, that is, acceleration = 0 when at terminal velocity ......

4. Find the position function s(t) of the object. Mr F says: Once you have v = v(t), solve dx/dt = v(t) ......

..
• Feb 9th 2008, 02:33 PM
mr fantastic
Quote:

Originally Posted by dark_knight_307
Could you show me how to do it, really fast like?

Read this. In fact, the very problem you have is done as an example.

• Feb 9th 2008, 02:36 PM
mr fantastic
Quote:

Originally Posted by Krizalid
Do you know what integration by parts technique is?

How is then you're covering ODE if you don't know how to integrate by parts?

Hmmmmmmmm ..........
• Feb 9th 2008, 02:39 PM
Krizalid
Kinda weird, huh?

It's like you just learnt something for a while.
• Feb 9th 2008, 02:40 PM
dark_knight_307
Quote:

Originally Posted by mr fantastic
Read this. In fact, the very problem you have is done as an example.

Is there any other way to integrate that?
• Feb 9th 2008, 02:42 PM
Krizalid
As I said before, you may perform a solution with complex numbers.

$\int {e^x \cos x\,dx} = \text{Re} \int {e^{(1 + i)x} \,dx} .$

Then you just solve that, take real parts and done.
• Feb 9th 2008, 02:46 PM
dark_knight_307
Quote:

Originally Posted by Krizalid
As I said before, you may perform a solution with complex numbers.

$\int {e^x \cos x\,dx} = \text{Re} \int {e^{(1 + i)x} \,dx} .$

Then you just solve that, take real parts and done.

Well I've never heard of that before, so....
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