My Math Forum  

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

Differential Equations Ordinary and Partial Differential Equations Math Forum


Reply
 
LinkBack Thread Tools Display Modes
October 4th, 2018, 06:07 AM   #1
Senior Member
 
Joined: Dec 2015
From: Earth

Posts: 248
Thanks: 27

Homogeneous solution explain

Given equation $\displaystyle y'+py=q$
for $\displaystyle q=0$ then $\displaystyle y=y_h$ is the homogeneous solution
(1) Explain why solution to equation is $\displaystyle y=y_h+y_p$
where $\displaystyle y_h$ - homogeneous and $\displaystyle y_p$ -particular
(2) Explain how to derive $\displaystyle y_p $ from $\displaystyle y_h$
How to derive $\displaystyle y=y_h + y_p$ ?
idontknow is offline  
 
October 4th, 2018, 11:11 AM   #2
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,445
Thanks: 2499

Math Focus: Mainly analysis and algebra
1) Suppose that $y_p$ is any solution of $y'+py=q$ and that $y_c$ is any solution of $y'+py=0$. Then if $y= y_c+y_p$, we have $y'=y_c'+y_p'$ and so \begin{align}y'+py &= (y_c'+y_p') + p(y_c+y_p) \\ &= (y_c' + py_c) + (y_p' + py_p) &(\text{just grouping the terms differently}) \\ &= 0 + q \\ &= q\end{align}
v8archie is offline  
October 4th, 2018, 12:03 PM   #3
SDK
Senior Member
 
Joined: Sep 2016
From: USA

Posts: 473
Thanks: 262

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
Originally Posted by v8archie View Post
1) Suppose that $y_p$ is any solution of $y'+py=q$ and that $y_c$ is any solution of $y'+py=0$. Then if $y= y_c+y_p$, we have $y'=y_c'+y_p'$ and so \begin{align}y'+py &= (y_c'+y_p') + p(y_c+y_p) \\ &= (y_c' + py_c) + (y_p' + py_p) &(\text{just grouping the terms differently}) \\ &= 0 + q \\ &= q\end{align}
This is true but only gets you one direction. He/she still has to show that if $x$ is any solution to this ODE, then it has the required form. For this it suffices to show that the homogenous equation has a unique solution (via another application of v8Archie's argument).

For this the usual approach is to assume write the equation as $x' = Ax$ so that $\exp(tA)$ is a solution. Now assume that $x$ is a solution and and show that $\exp(tA)x(t)$ is constant.

For the second part, this is just the variation of constants formula:
\[x(t) = \exp(tA) \left( \exp(-t_0A)x_0 + \int_{t_0}^{t} \exp(-sA) g(s) \ ds \right) \]
SDK is offline  
Reply

  My Math Forum > College Math Forum > Differential Equations

Tags
explain, homogeneous, solution



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Non-trivial solution of a homogeneous system jones123 Algebra 11 June 21st, 2016 01:22 AM
Particular Solution for homogeneous case JohnofGaunt Differential Equations 2 June 7th, 2014 06:42 AM
help need for finding non-homogeneous solution skvashok Applied Math 1 December 21st, 2012 06:05 AM
find particular solution of non-homogeneous mbradar2 Calculus 3 October 13th, 2010 08:49 PM
Non-trivial solution of a homogeneous system jones123 Calculus 2 December 31st, 1969 04:00 PM





Copyright © 2018 My Math Forum. All rights reserved.