My Math Forum  

Go Back   My Math Forum > College Math Forum > Differential Equations

Differential Equations Ordinary and Partial Differential Equations Math Forum


Thanks Tree8Thanks
  • 1 Post By idontknow
  • 2 Post By DarnItJimImAnEngineer
  • 2 Post By v8archie
  • 3 Post By v8archie
Reply
 
LinkBack Thread Tools Display Modes
October 22nd, 2019, 04:29 AM   #1
Senior Member
 
Joined: Dec 2015
From: Earth

Posts: 827
Thanks: 113

Math Focus: Elementary Math
Non-order DE

$\displaystyle y^{(4)} -y+e^{-x} =0$.
idontknow is offline  
 
October 22nd, 2019, 06:35 AM   #2
Senior Member
 
Joined: Dec 2015
From: Earth

Posts: 827
Thanks: 113

Math Focus: Elementary Math
One solution can be found below :
$\displaystyle (y'+y)^{(3)} -(y''-y')' -e^{x} (y''e^{-x} -e^{-x}y)=-e^{-x}$.
$\displaystyle [e^{-x}(ye^{x})']^{(3)}-(y''-y')'-e^{x}\frac{dW(e^{-x} , y)}{dx}=-e^{-x}$.
$\displaystyle \int d[e^{-x}(ye^{x})']''-\int d(y''-y')-\int e^{x}dW(e^{-x},y) =C+e^{-x}$.
$\displaystyle [e^{-x}(ye^x )']''-y''+y'-[e^{x}W(e^{-x},y)-\int W(e^{-x},y)e^{x}dx]=C+e^{-x}$.
$\displaystyle [e^{-x}(ye^x )']''-y''+\int ydx=C+e^{-x}$.
$\displaystyle -e^{-x} +y'''-[e^{-x}(ye^x )']'''=y$.
$\displaystyle -e^{-x} +y'''-[y'+y]'''=-e^{-x}-y'=y$.
$\displaystyle e^{x}(y'+y)=\int dye^{x} =-\int dx=C-x$.
$\displaystyle y=Ce^{-x}-xe^{-x}$. C-only one value.
Thanks from topsquark
idontknow is offline  
October 22nd, 2019, 08:02 AM   #3
Senior Member
 
Joined: Jun 2019
From: USA

Posts: 380
Thanks: 205

If it's a 4th-order ODE, the general solution should have four constants to solve for.
Why don't you just start with $y \sim e^{\lambda x}$?
Thanks from topsquark and idontknow
DarnItJimImAnEngineer is offline  
October 22nd, 2019, 08:17 AM   #4
Senior Member
 
Joined: Dec 2015
From: Earth

Posts: 827
Thanks: 113

Math Focus: Elementary Math
$\displaystyle \lambda^{4} e^{\lambda x} -e^{\lambda x} +e^{-x} \approx 0$;
How to continue this way ?
idontknow is offline  
October 22nd, 2019, 08:33 AM   #5
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,697
Thanks: 2681

Math Focus: Mainly analysis and algebra
4th order ode with constant coefficients
$$y''''-y=-e^{-x}$$
has characteristic equation (for the associated homogeneous equation)
\begin{align}
r^4-1 &= 0 \\
(r-1)(r+1)(r-i)(r+i) &= 0
\end{align}
So the complementary solutions are $$y_c = Ae^x + Be^{-x} + C\sin{x} + D\cos{x}$$
Now you just need the particular solution which should be easy to determine by the method of undetermined coefficients.
Thanks from topsquark and idontknow
v8archie is offline  
October 22nd, 2019, 08:36 AM   #6
Senior Member
 
Joined: Dec 2015
From: Earth

Posts: 827
Thanks: 113

Math Focus: Elementary Math
Quote:
Originally Posted by v8archie View Post
4th order ode with constant coefficients
$$y''''-y=-e^{-x}$$
has characteristic equation (for the associated homogeneous equation)
\begin{align}
r^4-1 &= 0 \\
(r-1)(r+1)(r-i)(r+i) &= 0
\end{align}
Seems like I'm nowhere at all; can you post the solution?

Last edited by skipjack; October 22nd, 2019 at 11:06 AM.
idontknow is offline  
October 23rd, 2019, 06:52 AM   #7
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,697
Thanks: 2681

Math Focus: Mainly analysis and algebra
The solution of any linear ODE with constant coefficients $$a_ny^{n} + a_{n-1}y^{n-1} + \ldots + a_1y' + a_0y =g(x)$$ is given by $$y = y_c + y_p$$ where $y_c$, the complimentary solution, is the general solution of the homogeneous equation $$a_ny^{n} + a_{n-1}y^{n-1} + \ldots + a_1y' + a_0y=0$$ and $y_p$, the particular solution, is any solution of the original equation.

The solutions of the homogeneous equation are all of the form $e^{rx}$ (where $r$ is possibly complex) and by substituting $y=e^{rx}$ into the homogenous equation we find the characteristic polynomial which can be solved for values of $r$, the number of solutions determined by the degree of the characteristic polynomial which in turn is determined by the order of the original ODE.

In this case we get \begin{align}
r^4-1 &= 0 \\
(r-1)(r+1)(r-i)(r+i) &= 0
\end{align}
It can be shown that real-valued solutions of the ODE with complex-conjugate values of $r = u \pm iv$ are equivalent to solutions $y = e^{ux}(A\sin vx + B\cos vx)$. Thus the complementary solution of $$y''''-y=-e^{-x}$$ is $$y_c = Ae^x + Be^{-x} + C\sin{x} + D\cos{x}$$

For the particular solution, using the method of undetermined coefficients (other methods are available) we guess the appropriate template for a solution, in this case $y_p = Exe^{-x}$ where $E$ is a constant to be determined. The $x$ in this term comes in because $Be^{-x}$ is already a solution (where $B$ is an arbitrary constant of integration).

Thus we substitute $y_p=Exe^{-x}$ into the original ODE to get
\begin{align}
Exe^{-x} - 4Ee^{-x} - Exe^{-x} &= -e^{-x} \\
\implies E &= \tfrac14
\end{align}

And so we have a solution \begin{align}
y &= y_c + y_p \\
&= \tfrac14xe^{-x} + Ae^x + Be^{-x} + C\sin{x} + D\cos{x}
\end{align}

All of the above is a method standard for solving a second order ODE with constant coefficients extended to the fourth order equation you presented. This extension should be taught in the same course as the method for solving second order equations.

Last edited by v8archie; October 23rd, 2019 at 07:00 AM.
v8archie is offline  
Reply

  My Math Forum > College Math Forum > Differential Equations

Tags
nonorder



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Rewriting second order to 1st order diff eqn Don96 Differential Equations 4 April 22nd, 2016 08:14 PM
Reduction of 2nd order PDE to a first order equations system math2016 Differential Equations 2 October 2nd, 2015 09:04 AM
Proof that group of order 2k containts order k subgroup Kappie Abstract Algebra 0 April 22nd, 2012 02:52 PM
Transform 2nd order ODE to two 1st order ODE using matrices Norm850 Calculus 2 March 7th, 2012 05:08 PM
If a has order hk modulo n, then a^h has order k mod n. Jamers328 Number Theory 1 December 2nd, 2007 09:21 PM





Copyright © 2019 My Math Forum. All rights reserved.