First order differential equations

In summary, the first differential equation has a solution that approaches zero as t→∞ when a particular condition is met on the parameter a.
  • #1
DmytriE
78
0

Homework Statement


Consider the first order differential equation

[itex]\frac{dx(t)}{dt} + ax(t) = f(t), x(0) = x_{0}, t\geq0[/itex]​

Suppose the "input signal" [itex] f(t)=e^{-t}, t\geq0 [/itex]. (a) Find the solution to the equation. Find a condition on the parameter a so that the solution of the (forced) system approaches zero as t→∞.

Homework Equations


[itex]\frac{dy}{dt} + p(x)y = 0 [/itex]

The Attempt at a Solution



Setup as a homogenous equation therefore f(t) = 0.
[itex]\frac{dx(t)}{dt} + ax(t) = 0 [/itex]

[itex]\frac{dx(t)}{x(t)} = -a*dt [/itex]

[itex] ln(x(t)) = -at [/itex]

[itex] x(t) = e^{-at} [/itex]

I don't know how to proceed any further...
 
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  • #2
What you found satisfies the homogeneous equation. You are still missing an integration constant.
Now you need one particular solution that satisfies the inhomeogeneous equation.
A times homogeneous solution + particular solution is ' the ' general solution
 
  • #3
Is the integration constant [itex] e^{\int x(t) dt} [/itex] or [itex] e^{\int a dt} [/itex]?

I guess the only integration factor that makes sense is the later of the two.
 
Last edited:
  • #4
No, I mean [itex] x(t) = Ce^{-at} [/itex] also satisfies the homogeneous equation.
 
  • #5
Next thing to try: [itex] C = C(t) [/itex]. Put this in the inhomogeneous equation and see what equation you get for C.
 
  • #6
I'm sorry but you've now lost me. Why am I trying C=C(t)?
 
  • #7
Well, the [itex] e^{-at} [/itex] part of [itex] x(t) = C(t)e^{-at} [/itex] helps you get rid of the [itex] +a x(t) [/itex] term. In return, you are left with a differential equation for [itex] C(t) [/itex] that you may be able to solve.
 
  • #8
DmytriE said:

Homework Statement


Consider the first order differential equation

[itex]\frac{dx(t)}{dt} + ax(t) = f(t), x(0) = x_{0}, t\geq0[/itex]​

Suppose the "input signal" [itex] f(t)=e^{-t}, t\geq0 [/itex]. (a) Find the solution to the equation. Find a condition on the parameter a so that the solution of the (forced) system approaches zero as t→∞.

Homework Equations


[itex]\frac{dy}{dt} + p(x)y = 0 [/itex]


The Attempt at a Solution



Setup as a homogenous equation therefore f(t) = 0.
[itex]\frac{dx(t)}{dt} + ax(t) = 0 [/itex]

[itex]\frac{dx(t)}{x(t)} = -a*dt [/itex]

[itex] ln(x(t)) = -at [/itex]

[itex] x(t) = e^{-at} [/itex]

I don't know how to proceed any further...

Either use Method of undetermined coefficients or use the integrating factor [itex]e^{at}[/itex].
 
  • #10
How can you tell when a first differential needs to use an integrating factor?
 
  • #11
DmytriE said:
How can you tell when a first differential needs to use an integrating factor?

The differential equation you have presented is a linear differential equation of first order and this immediately tells us to use the integrating factor.
 
  • #12
DmytriE said:
How can you tell when a first differential needs to use an integrating factor?

In this case, you don't need an integrating factor. Just put ##x(t)=Ce^{-t}## into the equation and solve for C. You can't always do that but it's a nice shortcut here. Can you see why you could guess it would work? In general, guessing a trial solution is often a good option. When you can.
 
  • #13
DmytriE said:
How can you tell when a first differential needs to use an integrating factor?
Look at the left side of your equation. "a" is a constant, so it is certainly going to integrate nicely. If "a" can be integrated easily with respect to t, that's your first indicator. Now look at the right side of the equation. It's a function only of t. So, when you multiply both sides by the integrating factor, the right side will still only be a function only of t (that can be integrated). That's your second indicator.

Chet
 
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  • #14
Perhaps it's good to give a general derivation of the integrating factor for 1st-order linear ODEs. The standard form of such an ODE is
[tex]y'(x)+p(x) y(x)=q(x),[/tex]
where [itex]p[/itex] and [itex]q[/itex] are given functions and the equation is to besolved for [itex]y(x)[/itex]. We look for the general solution.

The equation would be much easier, if we had a total derivative on the left-hand side, but that's not too difficult to achieve. You just substitute
[tex]z(x)=v(x) y(x).[/tex]
From the product rule you get
[tex]z'(x)=v(x) y'(x)+v'(x) y(x).[/tex]
Multiplying the ODE with [itex]v(x)[/itex] gives
[tex]v(x) y'(x)+p(x) v(x) y(x)=q(x) v(x).[/tex]
To make the left-hand side equal to [itex]z'(x)[/itex] you obviously have to determine [itex]v(x)[/itex] so that
[tex]v'(x)=p(x) v(x)[/tex]
This is very easily solved, because you can write
[tex]\frac{\mathrm{d}}{\mathrm{d} x} \ln \left (\frac{v(x)}{v_0} \right )=p(x).[/tex]
This gives
[tex]v(x)=v_0 \exp \left (\int_{0}^{x} \mathrm{d} x' p(x') \right ).[/tex]
Here [itex]v_0 \neq 0[/itex] is an arbitrary integration constant.

Thus, choosing [itex]v(x)[/itex] in this way we can write our ODE as
[tex]z'(x)=q(x) v(x)[/tex]
with the general solution
[tex]z(x)=z_0 + \int_0^{x} \mathrm{d} x' q(x') v(x').[/tex]
Finally we have
[tex]y(x)=\frac{z(x)}{v(x)}=\frac{z_0}{v(x)}+\frac{1}{v(x)} \int_0^{x} \mathrm{d} x' q(x') v(x').[/tex]
As you see, the choice of [itex]v_0[/itex] doesn't affect your solution in any way, because it only appears as an overall factor in front of the integration constant [itex]z_0[/itex], and this can be lumped into the overall integration constant of the general solution of the homogeneous equation. Finally we have the general solution of the ODE
[tex]y(x)=\exp \left (-\int_0^{x} \mathrm{d} x' p(x') \right ) \left [y_0+ \int_0^x \mathrm{d} x' q(x') \exp \left ( \int_0^x \mathrm{d} x'' p(x'') \right ) \right].[/tex]
Here [itex]y_0[/itex] is given by the initial-value condition [itex]y(0)=y_0[/itex].
 
  • #15
Thank you everyone! I finally figured it out using integrating factors. As for guessing, would you guess Ce^-t as a good solution because the function is forced by e^-t?
 
  • #16
DmytriE said:
Thank you everyone! I finally figured it out using integrating factors. As for guessing, would you guess Ce^-t as a good solution because the function is forced by e^-t?

It's a good guess because of the other side of the equation, if you differentiate ##e^{-t}## or multiply it by ##a##, you just get back multiples of ##e^{-t}##.
 

1. What is a first order differential equation?

A first order differential equation is an equation that involves an unknown function and its derivative. It is called a first order equation because it contains the first derivative of the unknown function, as opposed to higher derivatives.

2. How do you solve a first order differential equation?

There are several methods for solving first order differential equations, such as separation of variables, integrating factors, and using substitution. The method used will depend on the specific equation and its characteristics.

3. What are the applications of first order differential equations?

First order differential equations have numerous applications in various fields, such as physics, engineering, economics, and biology. They are used to model and analyze a wide range of phenomena, including growth and decay, motion, and population dynamics.

4. Are there any real-life examples of first order differential equations?

Yes, there are many real-life examples of first order differential equations. For instance, the rate of change of temperature in a room can be modeled using a first order differential equation. Another example is the growth of a population over time, which can also be described by a first order equation.

5. Can all first order differential equations be solved analytically?

No, not all first order differential equations can be solved analytically. Some equations may require numerical methods to find approximate solutions. However, there are certain types of first order equations, such as separable equations, that can be solved analytically in most cases.

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