Bit hazy on reduction of order

In summary, the conversation discusses the equation vt''-3v' = 0 and the process of solving it through reduction of order. The general solution is given as v(t)=C_1t^4+C_2, and the conversation also touches on the issue of treating second derivatives and the necessity of another equation for an explicit solution.
  • #1
Matt Jacques
81
0
Although on the whole my Boyce & DiPrima book is fairly good in of itself and compared with others, but of-coarse some areas are lacking.

vt'' - 3v' = 0

I suppose I integrate twice, but then what?
 
Physics news on Phys.org
  • #2
vt''-3v'? You might want to check that, v and t are both functions of some paramater?
 
  • #3
More specifically, v(t)*t'' - 3v(t) = 0

I'd go insane if I kept that (t) in there while differentiating and combining. :P
 
  • #4
And that factorizes as:

v(t)*(t''-3)=0

not sure that's right either as now the second v is not primed.
 
  • #5
v''(t)*t - 3v'(t) = 0

Sorry, this is it now.
 
  • #6
Well, one solution is v(t)=t^3, don't you have to use this to get the other solution?


Edit, of course it isn't. There's the constant solution, duh, and that gives the answer in the next post.
 
Last edited:
  • #7
The general solution

Hi;
The general solution is:
[tex]v(t)=C_1t^4+C_2[/tex]
Good luck,
Max.
 
  • #8
add the 3v' to the other side getting:
vt''=3v'
divide v to the other side and then integrate with respect to t and v, based on the side of the equation, being sure to keep the integration constants in.
 
  • #9
bowzerman said:
add the 3v' to the other side getting:
vt''=3v'
divide v to the other side and then integrate with respect to t and v, based on the side of the equation, being sure to keep the integration constants in.
No, you cannot treat second derivatives like that. If you really had a single equation in two dependent variables, you will need another equation to get an explicite solutions just as you cannot solve one numeric equation in two variables.

The problem is, after editing, tv"j- 3v'= 0.
Matt Jacques, since you explicitely refer to "reduction of order", I think you mean this: set u= v' so that u'= v" and you have the first order equation tu'- 3u= 0. Now, you can do exactly what bowzerman suggested, since this is now a first order equation.
t du/dt= 3u so du/u= 3t dt. ln|u|= (3/2)t2+ C and then
[tex]u= C_1 e^{(3/2)t^2[itex].
Since u= dv/dt, we have
[tex]dv= C_1 e^{3/2}t^2 dt[/tex]
and only have to integerate again. I'll leave that to you!
 
Last edited by a moderator:
  • #10
bowzerman said:
add the 3v' to the other side getting:
vt''=3v'
divide v to the other side and then integrate with respect to t and v, based on the side of the equation, being sure to keep the integration constants in.
No, you cannot treat second derivatives like that. If you really had a single equation in two dependent variables, you will need another equation to get an explicite solutions just as you cannot solve one numeric equation in two variables.

The problem is, after editing, tv"- 3v'= 0.
Matt Jacques, since you explicitely refer to "reduction of order", I think you mean this: set u= v' so that u'= v" and you have the first order equation tu'- 3u= 0. Now, you can do exactly what bowzerman suggested, since this is now a first order equation.
t du/dt= 3u so du/u= 3t dt. ln|u|= (3/2)t2+ C and then
[tex]u= C_1 e^{(3/2)t^2[tex].
Since u= dv/dt, we have
[tex]dv= C_1 e^{3/2}t^2 dt[/tex]
and only have to integerate again. I'll leave that to you!
 
  • #11
yeah, sorry, I was just wrote that one too quick. Forgot to clarify.
 
  • #12
oh, and shouldn't that be:
t du/dt = 3u
1/u du = 3/t ?
ln(u)=3ln(t) +C
u= c*t^3

, then sub, integrate, and then plug in?
 
Last edited:

1. What is the concept of reduction of order in scientific terms?

Reduction of order is a mathematical method used in solving differential equations. It involves reducing a higher order differential equation to a first order differential equation by substituting a new variable.

2. Why is reduction of order important in scientific research?

Reduction of order is important because it simplifies complex mathematical problems and makes them easier to solve. It also helps in finding general solutions to differential equations, which are used in various scientific fields such as physics and engineering.

3. What are the steps involved in reduction of order?

The first step is to identify the highest order derivative in the differential equation. Then, a new variable is substituted to represent the derivative of the highest order. This reduces the original equation to a first order equation that can be solved using standard techniques.

4. How does reduction of order relate to real-life applications?

Reduction of order is used in various real-life applications, such as predicting the growth of bacteria or analyzing the behavior of electric circuits. It is also used in solving problems in mechanics, heat transfer, and fluid dynamics.

5. Are there any limitations to using reduction of order in scientific research?

One limitation is that not all differential equations can be reduced to first order equations using this method. Additionally, the solutions obtained may not always be accurate, especially in cases where the original equation is highly nonlinear.

Similar threads

  • Differential Equations
Replies
3
Views
2K
  • Differential Equations
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
Replies
1
Views
1K
  • Science and Math Textbooks
Replies
15
Views
2K
  • Atomic and Condensed Matter
Replies
5
Views
1K
  • Differential Equations
Replies
10
Views
2K
  • STEM Academic Advising
2
Replies
60
Views
3K
  • Differential Equations
Replies
3
Views
2K
Replies
1
Views
1K
Back
Top