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May 14th, 2018, 05:58 AM   #1
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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
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 May 14th, 2018, 06:31 AM #2 Senior Member   Joined: Oct 2009 Posts: 519 Thanks: 165 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: 7,405 Thanks: 2477 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
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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: 19,546 Thanks: 1754 No.
 May 16th, 2018, 03:03 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 The characteristic equation for 43 is $r^3- 4r^2+ r= r(r^2- 4r+ 1)= 0$. r= 0 is a root. $r^2- 4r+ 1= r^2- 4r+ 4- 4+ 1= (r+ 2)^2- 3= 0$ so that $r= -2+\sqrt{3}$ and [tex]r= -2- \sqrt{3}[tex] are the other 2 roots. The general solution to the differential equation is $y= e^{-2x}(C_1e^{-x\sqrt{3}}+ C_2e^{-x\sqrt{3}}+ C_3$. 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 $-sin^2(x+ c)- cos^2(x+ c)+ 1= 0$. 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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