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
June 1st, 2017, 06:42 AM   #1
pkm
Newbie
 
Joined: Jun 2017
From: South Africa

Posts: 1
Thanks: 0

Ode

Hi everyone, I have been trying to solve this equation, but to no avail. Can you help me?

y''+y'/x+y/x^2=0

Last edited by skipjack; June 1st, 2017 at 08:12 AM.
pkm is offline  
 
June 1st, 2017, 08:11 AM   #2
Global Moderator
 
Joined: Dec 2006

Posts: 18,154
Thanks: 1418

Let's assume that x > 0. Substitute x = e^u. The resulting equation is easy to solve.
skipjack is online now  
June 15th, 2017, 06:18 AM   #3
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 2,822
Thanks: 750

Not with those "x"s there, it isn't. The first thing I would do is multiply by $\displaystyle x^2$ so we have $\displaystyle x^2y''+ xy'+ y= 0$, an "Euler type" or "equi-potential" equation.

Now, try something of the form $\displaystyle y= x^s$. Then $\displaystyle y'= sx^{s-1}$ and $\displaystyle y''= s(s- 1)x^{s- 1}$ so the equation becomes $\displaystyle s(s- 1)x^s+ sx^s+ x^s= (s(s- 1)+ s+ 1)x^s= 0$. $\displaystyle x^s$ is not always 0 so we must have $\displaystyle s(s- 1)+ s+ 1= s^2+ 1= 0$ which has roots s= i and s= -i. So the general solution to the differential equation is $\displaystyle = Ax^i+ Bx^{-i}$. To get that in terms of real numbers only, write $\displaystyle x^i= e^{\ln(x^i)}= e^{i\ln(x)}$ and use the fact that $\displaystyle e^{ia}= \cos(a)+ i\sin(a)$ to get $\displaystyle x^{i}= \cos(\ln(x))+ i \sin(\ln(x))$. Similarly, $\displaystyle x^{-i}= \cos(-\ln(x))+ i \sin(-\ln(x))= \cos(\ln(x))- \sin(\ln(x))$. We can absorb the "i" into the constants, A and B, and write the general solution as $\displaystyle y= A\cos(\ln(x))+ B \sin(\ln(x))$.

(Another way to approach this is to use the fact that the substitution t= ln(x) reduces it to an equivalent equation with constant coefficients. If t = ln(x), then $\displaystyle \frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}$ and $\displaystyle \frac{d^2y}{dx^2}= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x^2}\frac{dy}{dt}$ so that the differential equation becomes $\displaystyle \frac{d^2y}{dt^2}+ y= 0$.)

Last edited by skipjack; June 15th, 2017 at 06:39 AM.
Country Boy is offline  
June 15th, 2017, 06:35 AM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 18,154
Thanks: 1418

Why disagree with substituting x = e^u, yet demonstrate how easily the substitution t = ln(x) works?
skipjack is online now  
Reply

  My Math Forum > College Math Forum > Differential Equations

Tags
ode



Thread Tools
Display Modes






Copyright © 2017 My Math Forum. All rights reserved.