Solving linear differential equations

In summary: Don't give up on it.In summary, the conversation discusses two linear differential equations in terms of the Laplace Operator and the process of solving them using integration by parts. The conversation also touches on the idea of equating the Laplace transforms of two equal linear equations. The speaker struggles with the integration process and decides to leave the Laplace transform for now.
  • #1
tomwilliam2
117
2
I have the following linear differential equations:
##A\dot{x} + By = 0##
##C\dot{y} + Dx = 0##
Where x and y are functions of t, and A through D are constants.
I can solve this fairly easily by differentiating the first equation, rearranging, and removing one of the variables, which gives me a fairly straightforward to solve 2nd order differential equation.
However, my textbook says "these are linear differential equations, whose characteristic equation, in terms of the Laplace Operator, is..." and goes on to produce the same characteristic equation as I get. I understand that when the linear equations get more complicated, you have to use a Laplace transform to solve them. So I thought I'd give it a go, but can't get the same characteristic equation. I'm not sure where I'm going wrong, so if I put my working out here, could someone point me to the mistake(s)?
I'm starting with the first equation:
## F(s) = A \int_{0}^{\infty} \dot{x}e^{-st} dt + B \int_{0}^{\infty} ye^{-st}dt##
Using integration by parts:
##F(s) = A\left(xe^{-st} + s \int_{0}^{\infty}xe^{-st} dt \right) + B/s \left( \int_{0}^{\infty}e^{-st} dt - ye^{-st}\right)##
Now applying integration by parts again:
##F(s) = A\left((xe^{-st} -xe^{-st} - 1/s\right) -By/s - B/s^2 \left(\int_{0}^{\infty}e^{-st} dt\right)##
And I think here I have to apply the limits to clear it up, so:
##F(s) = -A/s - By/s + B/s^2 = \frac{B(1-ys)}{s^2} - \frac{A}{s}##
But I think this has already gone wrong somewhere, as I don't see this becoming a quadratic equation.
Presuming I can fix this integration by parts, what is the next step? To do the same thing with the other linear equation and then equate them?
Thanks in advance for any suggestions.
 
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  • #2
I don't see where some terms of the second line come from. ##xe^{-st}## needs the be evaluated at the boundary points ##0## and ##\infty##, since that's how partial integration works. In the ##B## term, why did the ##y## disappear? The only think I can think of that I'm missing is that you may have used the differential equations.
 
  • #3
Lucas SV said:
I don't see where some terms of the second line come from. ##xe^{-st}## needs the be evaluated at the boundary points ##0## and ##\infty##, since that's how partial integration works. In the ##B## term, why did the ##y## disappear? The only think I can think of that I'm missing is that you may have used the differential equations.
Thanks for your comment. I missed out some of the steps, as it was painstaking to write out, but I might have made a mistake somewhere. If I take just the B term:
##F_B(s) = B \int_{0}^{\infty} ye^{-st} dt##
Using integration by parts, we set
##v = y##
##v' = 1##
##u = -\frac{1}{s}e^{-st}##
##u' = e^{-st}##
and ##\int u'v = uv - \int uv'##
So
##B \int_{0}^{\infty} ye^{-st} dt = B \left(-\frac{y}{s}e^{-st} + \frac{1}{s}\int_{0}^{\infty}e^{-st}\right) dt## evaluated between t=0 and t=infinity.
So:
##F_B(s) = B\left(-\frac{y}{s}e^{-st} + \frac{1}{s^2}e^{-st}\right)## from t=0 to t=infinity.
##F_B(s) = \frac{By}{s} - \frac{B}{s^2}##
So that's how I got the result for the B term. Did I make a mistake?
 
  • #4
Yes you did, your integration variable is ##t## and ##y## is a function of ##t## as stated in the beggining. So ##v'=y'## is only equal to ##1## if ##y## is a linear function with respect to ##t##.
 
  • #5
Lucas SV said:
Yes you did, your integration variable is ##t## and ##y## is a function of ##t## as stated in the beggining. So ##v'=y'## is only equal to ##1## if ##y## is a linear function with respect to ##t##.
Ah, of course. How silly of me. So I'll go back and try to solve it again. I presume that when I get an expression for the Laplace transform of the first equation, I have to then equate it with the transform of the second equation?
That makes me wonder, if f(x) = g(x), then does L(f) = L(g)? I.e. if two linear equations are equal to each other, then are their laplace transforms also equal to each other?
Thanks again.
 
Last edited:
  • #6
tomwilliam2 said:
That makes me wonder, if f(x) = g(x), then does L(f) = L(g)? I.e. if two linear equations are equal to each other, then are their laplace transforms also equal to each other?
Yes this is true. If any two objects in mathematics are equal, then you can substitute one for the other in an arbitrary expression. To be more precise you have to say ##f(x)=g(x)## for all ##x\in D##, where ##D## is the common domain of ##f## and ##g## (they need to have common domain in order to be equal).
 
  • #7
Thanks. So I'm left trying to integrate:
##\int_{0}^{\infty}\dot{x}e^{-st}dt##
Which has got me completely stuck.
I think I'll just leave the LaPlace transform. It's a shame though, because I understand it is a good tool to use in some cases where the differential equations are more difficult than here.
 
  • #8
tomwilliam2 said:
Thanks. So I'm left trying to integrate:
##\int_{0}^{\infty}\dot{x}e^{-st}dt##
Which has got me completely stuck.
I think I'll just leave the LaPlace transform. It's a shame though, because I understand it is a good tool to use in some cases where the differential equations are more difficult than here.
It's fine, come back at a later stage. Sometimes there are things we do not see straight away when learning, but some time later when we come back to it, it just clicks and it all makes sense.
 

1. What is a linear differential equation?

A linear differential equation is an equation that involves a dependent variable, its derivatives, and possibly independent variables, with the highest derivative having a power of 1. It can be written as a first-order or higher-order equation.

2. How do you solve a linear differential equation?

To solve a linear differential equation, you must first identify the type of equation (homogeneous, non-homogeneous, exact, etc.) and then use various techniques such as separation of variables, integrating factors, or the method of undetermined coefficients.

3. What is the difference between a homogeneous and non-homogeneous linear differential equation?

A homogeneous linear differential equation has all terms involving the dependent variable and its derivatives, while a non-homogeneous equation has additional terms that do not involve the dependent variable or its derivatives. The solution method for each type of equation is different.

4. Can you give an example of a linear differential equation?

One example of a linear differential equation is y' + 2xy = 3x, where y is the dependent variable and x is the independent variable. This is a first-order, non-homogeneous equation.

5. What are some real-life applications of solving linear differential equations?

Linear differential equations are used in various fields of science and engineering to model and predict the behavior of systems. Some examples include predicting population growth, analyzing electric circuits, and modeling chemical reactions.

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