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 Lalitha183 May 14th, 2018 05:58 AM

Differential Equations

1 Attachment(s)
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

 Micrm@ss May 14th, 2018 06:31 AM

Why don't you just plug the different solutions into the equation and see whether you get a correct result?

 v8archie May 14th, 2018 09:10 AM

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.

 Lalitha183 May 15th, 2018 09:18 PM

Quote:
 Originally Posted by Micrm@ss (Post 594340) 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 ?

 skipjack May 15th, 2018 11:07 PM

No.

 Country Boy May 16th, 2018 03:03 AM

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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