User Name Remember Me? Password

 Differential Equations Ordinary and Partial Differential Equations Math Forum

May 14th, 2018, 05:58 AM   #1
Senior Member

Joined: Nov 2015
From: hyderabad

Posts: 236
Thanks: 2

Differential Equations

I have attached the question images Q43 & Q44. These are my previous year questions, which I was unable to solve. Kindly help me.
Thank you
Attached Images 15263061357751559544980.jpg (89.7 KB, 24 views) May 14th, 2018, 06:31 AM #2 Senior Member   Joined: Oct 2009 Posts: 850 Thanks: 327 Why don't you just plug the different solutions into the equation and see whether you get a correct result? May 14th, 2018, 09:10 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra That's certainly the most direct approach. I think that the first question has an error. None of them look like solutions. I suspect the second also has an error, but it does at least have a solution. May 15th, 2018, 09:18 PM   #4
Senior Member

Joined: Nov 2015
From: hyderabad

Posts: 236
Thanks: 2

Quote:
 Originally Posted by Micrm@ss Why don't you just plug the different solutions into the equation and see whether you get a correct result?
Q43. I'm getting the general solution as c1e^(2+sqrt(3))x + c2e^(2-sqrt(3))x
But I'm not sure what else needs to be done to derive one of the solutions given.
Q44. I have substituted the solutions in the D.E and I'm getting option A & B as solutions. Is it correct ?

Last edited by Lalitha183; May 15th, 2018 at 09:59 PM. May 15th, 2018, 11:07 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,937 Thanks: 2210 No. May 16th, 2018, 03:03 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 The characteristic equation for 43 is . r= 0 is a root. so that and [tex]r= -2- \sqrt{3}[tex] are the other 2 roots. The general solution to the differential equation is . You forgot the constant solution. For 44 if you set y= x, then y'= 1 and y''= 0 so the equation becomes x(0)- 1+ 1= 0. If y= sin(x+ c) then y'= cos(x+ c) and y''= -sin(x+ c). The equation becomes . If y= sinh(x+ c) then y'= cosh(x+ c) and y''= sinh(x+ c). The equation becomes [tex]sinh^2(x+ c)- cosh^2(x+ c)+ 1= 0. Tags differential, equations, exam, question Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post hyperbola Calculus 34 November 26th, 2016 05:49 AM rishav.roy10 Differential Equations 0 August 21st, 2013 05:59 AM Felpudio Differential Equations 1 May 7th, 2013 06:20 AM suomik1988 Differential Equations 9 February 2nd, 2011 11:08 AM Matt Differential Equations 1 November 13th, 2009 01:34 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      