Math 2650 - Linear Differential EquationsA. J. MeirCopyright (C) A. J. Meir. All rights reserved.This worksheet is for educational use only. No part of this publication may be reproduced or transmitted for profit in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system without prior written permission from the author. Not for profit distribution of the software is allowed without prior written permission, providing that the worksheet is not modified in any way and full credit to the author is acknowledged.Separation of VariablesSeparation of variables is an analytical method tha may be used to solve certain d.e.s (and certain i.v.p.s). At the heart of the method is the chain rule.Consider a (first order) differential equation that may be written asNiMvKiYlI2R4RyIiIiUjZHRHISIiKiYtJSJmRzYjJSJ4R0YmLSUiZ0c2IyUidEdGJg==.That is, the derivative of the function is equal to the product of a function of only NiMlInhH and a function of only NiMlInRH (the r.h.s. of the equation).If NiMwLSUiZkc2IyUieEciIiE= we can writeNiMqJiIiIkYkLSUiZkc2IyUieEchIiI= NiMvKiYlI2R4RyIiIiUjZHRHISIiLSUiZ0c2IyUidEc=.Recall NiMvJSJ4R0Yk(NiMlInRH), so if we can find a function NiMtJSJIRzYjJSJ4Rw== with NiMvKiYlI2RIRyIiIiUjZHhHISIiKiZGJkYmLSUiZkc2IyUieEdGKA==, using the chain rule we have thatNiMqJiUiZEciIiIlI2R0RyEiIg==NiMvLSUiSEc2IyUieEcqJiUiZEciIiIlI2R4RyEiIg==NiMtJSJIRzYjJSJ4Rw==NiMqJiUjZHhHIiIiJSNkdEchIiI= = NiMqJiIiIkYkLSUiZkc2IyUieEchIiI= NiMqJiUjZHhHIiIiJSNkdEchIiI=soNiMqJiUiZEciIiIlI2R0RyEiIg==NiMvLSUiSEc2IyUieEctJSJnRzYjJSJ0Rw==.Integrating both sides of the equation w.r.t. NiMlInRH we getNiMvLSUiSEc2IyUieEcsJi0lJGludEc2JC0lImdHNiMlInRHRi8iIiIlImNHRjA=and sinceNiMvLSUiSEc2IyUieEctJSRpbnRHNiQqJiIiIkYsLSUiZkdGJiEiIkYn,the solution of the d.e. is given (implicitly) byNiMvLSUkaW50RzYkKiYiIiJGKC0lImZHNiMlInhHISIiRiwsJi1GJTYkLSUiZ0c2IyUidEdGNEYoJSJDR0Yo.Note: the above is an implicit solution. Also note the arbitrary constant NiMlIkNH, this is the general solution. If we are also given an initial condition we can use it to solve for NiMlIkNH and obtain a particular solution.Examples:1.NiMvKiYlI2R4RyIiIiUjZHRHISIiKiYlImtHRiYlInhHRiY=if NiMwJSJ4RyIiIQ==, seperating the variablesNiMqJiIiIkYkJSJ4RyEiIg== NiMlI2R4Rw=== NiMqJiUia0ciIiIlI2R0R0YlintegratingNiMvLSUjbG5HNiMtJSRhYnNHNiMlInhHLCYqJiUia0ciIiIlInRHRi5GLiUiQ0dGLg==exponentiatingNiMvLSUkYWJzRzYjJSJ4Ry0lJGV4cEc2IywmKiYlImtHIiIiJSJ0R0YuRi4lIkNHRi4=soNiMvLSUkYWJzRzYjJSJ4RyomJiUiQ0c2IyIiIkYsLSUkZXhwRzYjKiYlImtHRiwlInRHRixGLA== where NiMvJiUiQ0c2IyIiIi0lJGV4cEc2I0Yland absorbing the sign into the constant NiMmJSJDRzYjIiIi we get the solutionNiMvLSUieEc2IyUidEcqJiYlIkNHNiMiIiMiIiItJSRleHBHNiMqJiUia0dGLUYnRi1GLQ==another solution isNiMvLSUieEc2IyUidEciIiE=.What is the interval of existence of these solutions?Exercise: verify that the above are solutions of the differential equation.2.NiMvKiYlI2R4RyIiIiUjZHRHISIiKiYlInhHIiIjJSJ0R0Ymif NiMwJSJ4RyIiIQ==NiMvKiYlI2R4RyIiIiokJSJ4RyIiIyEiIiomJSJ0R0YmJSNkdEdGJg==a solution is given byNiMvLSUkaW50RzYkKiYiIiJGKCokJSJ4RyIiIyEiIkYqLCYtRiU2JCUidEdGMEYoJSJDR0Yorestart;int(1/x^2,x);

NiMsJComIiIiRiUlInhHISIiRic=

int(t,t);

NiMsJCokKSUidEciIiMiIiIjRihGJw==

So a solution isNiMvLCQqJiIiIkYmJSJ4RyEiIkYoLCYqJiUidEciIiNGLEYoRiYlIkNHRiY=.Solving for NiMlInhHsolve(-1/x = t^2/2+C,x);

NiMsJComIiIiRiUsJiokKSUidEciIiNGJUYlKiZGKkYlJSJDR0YlRiUhIiIhIiM=

we obtainNiMvLSUieEc2IyUidEcsJComIiIjIiIiLCYqJEYnRipGKyomRipGKyUiQ0dGK0YrISIiRjA=orNiMvLSUieEc2IyUidEcqJiIiIyIiIiwmJSJDR0YqKiRGJ0YpISIiRi4=.another solution isNiMvLSUieEc2IyUidEciIiE=.What is the interval of existence of these solutions? What does it depend on?Exercise: verify that the above are solutions of the differential equation.3.NiMvKiYlI2R4RyIiIiUjZHRHISIiLCYqJiUieEciIiMtJSRzaW5HNiMlInRHRiZGJiomRitGJkYtRiZGJg==rewrite the equation asNiMvKiYlI2R4RyIiIiUjZHRHISIiKiYsJiokJSJ4RyIiI0YmRixGJkYmLSUkc2luRzYjJSJ0R0Ymif NiMwLCYqJCUieEciIiMiIiJGJkYoIiIhNiMvKiYlI2R4RyIiIiwmKiQlInhHIiIjRiZGKUYmISIiKiYtJSRzaW5HNiMlInRHRiYlI2R0R0Yma solution is given byNiMvLSUkaW50RzYkKiYiIiJGKCwmKiQlInhHIiIjRihGK0YoISIiRissJi1GJTYkLSUkc2luRzYjJSJ0R0Y0RiglIkNHRig=int(1/(x^2+x),x);

NiMsJi0lI2xuRzYjJSJ4RyIiIi1GJTYjLCZGJ0YoRihGKCEiIg==

int(sin(t),t);

NiMsJC0lJGNvc0c2IyUidEchIiI=

So a solution isNiMvLCYtJSNsbkc2IyUieEciIiItRiY2IywmRihGKUYpRikhIiIsJi0lJGNvc0c2IyUidEdGLSUiQ0dGKQ==solve(ln(x)-ln(x+1) = -cos(t)+C,x);

NiMqJiIiIkYkLCYtJSRleHBHNiMsJi0lJGNvc0c2IyUidEdGJCUiQ0chIiJGJEYkRi9GLw==

soNiMvLSUieEc2IyUidEcqJiIiIkYpLCYqJiUiQ0dGKS0lJGV4cEc2Iy0lJGNvc0dGJkYpRilGKSEiIkYysolve(x^2+x=0);

NiQiIiEhIiI=

so additional solutions areNiMvLSUieEc2IyUidEciIiE= and NiMvLSUieEc2IyUidEcsJCIiIiEiIg==.What is the interval of existence of these solutions? What does it depend on?Exercise: verify that the above are solutions of the differential equation.Problem:Solve the equationNiMvKiYlI2R4RyIiIiUjZHRHISIiKiYlInhHRiYlInRHRig=.