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May 28th, 2017, 05:35 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Higher order D.E
The general solution of $y'''4y''+y'=0$ is A) $c_1 \sinh^2x+c_2\cosh^2x+c_3$ B) $c_1 \sinh2x+c_2 \cosh2x+c_3$ C) $c_1 \sin2x+c_2 \cos2x+c_3$ D) $c_1e^{2x}+c_2e^{2x}+c_3$ I got the roots as follows : $0$,$\displaystyle 2\pm\sqrt{3}$ How should I include these roots in the general solution format and which one is it? Please help! Thank you. Last edited by skipjack; May 28th, 2017 at 07:16 PM. 
May 28th, 2017, 09:08 AM  #2 
Member Joined: May 2017 From: Russia Posts: 33 Thanks: 4 
The solution is found in the form $\displaystyle C\cdot e^{\lambda x}, C\in \mathbb{R}$. One substitutes it for $\displaystyle y$: $\displaystyle (C\cdot e^{\lambda x})''' 4(C\cdot e^{\lambda x})''+ (C\cdot e^{\lambda x})' =0.$ $\displaystyle (\lambda^34\lambda^2+\lambda)C\cdot e^{\lambda x}=0.$ So, $\displaystyle y(x)=C_1\cdot e^{\lambda_1 x} +C_2\cdot e^{\lambda_2 x}+C_3\cdot e^{\lambda_3 x}.$ Last edited by ABVictor; May 28th, 2017 at 09:11 AM. 
May 28th, 2017, 12:26 PM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,822 Thanks: 750 
The characteristic equation is . We have r = 0 and so or or . That means that the general solution to the differential equation is . NONE of the given possibilities is correct. Last edited by skipjack; May 28th, 2017 at 07:46 PM. 
May 28th, 2017, 06:52 PM  #4  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Quote:
It is a Post graduation Entrance paper and moreover this question can have multiple answers! Please check if any other possible way to get the general solution Last edited by skipjack; May 28th, 2017 at 07:46 PM.  
May 28th, 2017, 07:23 PM  #5 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,408 Thanks: 480 Math Focus: Yet to find out.  
May 28th, 2017, 07:43 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,154 Thanks: 1418 
Obviously, "$t$" should be "$x$". Are you sure you typed the original equation correctly, Lalitha183?

May 28th, 2017, 08:09 PM  #7 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  
May 28th, 2017, 10:02 PM  #8 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  I have read it. But the option doesn't seems to be equivalent to the answer that came up.
Last edited by skipjack; May 28th, 2017 at 10:49 PM. 
May 28th, 2017, 10:12 PM  #9  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,408 Thanks: 480 Math Focus: Yet to find out.  Quote:
Last edited by skipjack; May 28th, 2017 at 10:49 PM.  

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