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May 14th, 2018, 05:58 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 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 
May 14th, 2018, 06:31 AM  #2 
Senior Member Joined: Oct 2009 Posts: 403 Thanks: 139 
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: 7,308 Thanks: 2443 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: 232 Thanks: 2  Quote:
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: 18,962 Thanks: 1606 
No.

May 16th, 2018, 03:03 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,165 Thanks: 867 
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. 

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