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 May 28th, 2017, 04:35 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 06:16 PM.
 May 28th, 2017, 08:08 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 $\displaystyle y$: $\displaystyle (C\cdot e^{\lambda x})''' -4(C\cdot e^{\lambda x})''+ (C\cdot e^{\lambda x})' =0.$ $\displaystyle (\lambda^3-4\lambda^2+\lambda)C\cdot e^{\lambda x}=0.$ 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 08:11 AM.
 May 28th, 2017, 11:26 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,647 Thanks: 680 The characteristic equation is $r^3- 4r^2+ r= r(r^2- 4r+ 1)= r(r^2- 4r+ 4- 4+ 1)= r(r^2- 4r+ 4- 3)= r((r- 2)^2- 3)= 0$. We have r = 0 and $(r- 2)^2= 3$ so $r= 0$ or $r= 2+ \sqrt{3}$ or $r= 2- \sqrt{3}$. That means that the general solution to the differential equation is $y(x)= A+ e^{2t}(Be^{t\sqrt{3}}+ Ce^{-t\sqrt{3}})$. NONE of the given possibilities is correct. Last edited by skipjack; May 28th, 2017 at 06:46 PM.
May 28th, 2017, 05:52 PM   #4
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Quote:
 Originally Posted by Country Boy The characteristic equation is $r^3- 4r^2+ r= r(r^2- 4r+ 1)= r(r^2- 4r+ 4- 4+ 1)= r(r^2- 4r+ 4- 3)= r((r- 2)^2- 3)= 0$. We have r = 0 and $(r- 2)^2= 3$ so $r= 0$ or $r= 2+ \sqrt{3}$ or $r= 2- \sqrt{3}$. That means that the general solution to the differential equation is $y(x)= A+ e^{2t}(Be^{t\sqrt{3}}+ Ce^{-t\sqrt{3}})$. NONE of the given possibilities is correct.
How could it be
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 06:46 PM.

May 28th, 2017, 06:23 PM   #5
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 Originally Posted by Lalitha183 How could it be 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

 May 28th, 2017, 06:43 PM #6 Global Moderator   Joined: Dec 2006 Posts: 17,914 Thanks: 1382 Obviously, "$t$" should be "$x$". Are you sure you typed the original equation correctly, Lalitha183?
May 28th, 2017, 07:09 PM   #7
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 Originally Posted by skipjack Obviously, "$t$" should be "$x$". Are you sure you typed the original equation correctly, Lalitha183?
Yes, I did.

I have read the post #2...So that could leave option D as the answer ?

May 28th, 2017, 09:02 PM   #8
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Quote:
 Originally Posted by Joppy Did you read post #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 09:49 PM.

May 28th, 2017, 09:12 PM   #9
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 Originally Posted by Lalitha183 I have read it. But the option doesn't seems to be equivalent to the answer that came up.
Take a look at the approximate form to the solution here.

Last edited by skipjack; May 28th, 2017 at 09:49 PM.

May 28th, 2017, 09:15 PM   #10
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Quote:
 Originally Posted by Joppy Take a look at the approximate form to the solution here.
That link is not working it seems.

Last edited by skipjack; May 28th, 2017 at 09:47 PM.

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