Find a linear homogeneous equation with given general solution

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Discussion Overview

The discussion revolves around finding a linear homogeneous constant-coefficient differential equation that corresponds to a given general solution. Participants explore various approaches to derive the differential equation from the provided solution, which includes exponential and polynomial terms.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant seeks assistance in deriving a linear homogeneous differential equation from the general solution y(x)=C1e^x+(C2+C3x+C4x^2)e-x.
  • Another participant proposes taking the fourth derivative of the solution but expresses uncertainty about the next steps.
  • A different participant suggests the differential equation -y''''-y''=0, but is unsure of its correctness.
  • It is noted that the characteristic equation should have one distinct real root and one repeated real root, leading to a factorization of (r-1)(r+1)^3.
  • One participant attempts to rewrite the factorization as (r-1)r^3 and derives the equation y''''-y'''=0, questioning its validity.
  • Another participant corrects the factorization, indicating that multiplying (r-1)(r+1)^{3} yields r^{4}+2r^{3}-2r-1.
  • Subsequently, a participant proposes the differential equation y''''-2y'''-2y-1=0, seeking confirmation.
  • Another participant presents the equation in a different form: y^{(4)}+2y^{(3)}-2y'-y, while clarifying the roots.
  • A new problem involving a second-order differential equation with imaginary roots is introduced, where a participant expresses difficulty in finding a particular solution for cos 4t+2 sin t.
  • One participant suggests a form for the particular solution as Yh=Acos4t +Bsin4t, while acknowledging the constant factor.
  • A participant reminds others that such problems should be posted in the Homework & Coursework section, leading to the closure of the thread.

Areas of Agreement / Disagreement

Participants express various approaches and solutions, but there is no consensus on the correct differential equation or the method to find the particular solution for the second-order equation. The discussion remains unresolved with multiple competing views.

Contextual Notes

Some participants' approaches depend on specific assumptions about the roots and factorizations, which may not be universally accepted. There are also unresolved steps in deriving particular solutions for the second-order differential equation.

pedro123
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I need help finding a linear homogenous constant-coefficient differential equation with the given general solution.

y(x)=C1e^x+(C2+C3x+C4x^2)e-x


2. I tried to come with differential equation but this is it
I can 't seem how to begin
 
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I came with taking the fourt derivate of y''''=Ce^-x(C3(x2-8x+12)+C2(x-4)+C+e^2x so Idont know what to do next. help please asap.
 
is the the differential equation -y''''-y''=0 I am not sure
 
The general solution implies that the characteristic equation has one distinct real root and one repeated real root, it can be factored as (r-1)(r+1)^3
 
so if its factored can be written as (r-1)r^3

and alternate fom beign r^4-r^3

making the general solution

y''''-y'''=0

is this alright thanks.
 
Not quite. If you multiply (r-1)(r+1)^{3}, you should get
(r-1)(r^{3}+3r^{2}+3r+1)=r^{4}+2r^{3}-2r-1
 
making the differential equation

y''''-2y'''-2y-1=0

is this alright and sorry about my mistake in factoring
 
The differential equation would be

y^{(4)}+2y^{(3)}-2y'-y

Remember that r=r^{1} and 1=r^{0}
 
thanks can help solve this problemI have a problem which in involves a second order differential equations with imaginary roots and I can seem to know how to finish the problem.

d^2y/dt^2 +15y =cos 4t+2 sin t

this is what I got so far r^2+15=0 for the homogeneous part

r=+-(√15)

Yh=C1cos√15+C2sin√15

and now the part that follows is to come with a particular solution for

cos 4t+2 sin t

but I don't know how to properlly set up
 
  • #10
and thanks for helping me so far
 
  • #11
I know that 2 is a constant but can my particular solution be

Yh=Acos4t +Bsin4t
 
  • #12
Problems such as these should be posted in the Homework & Coursework section, not in the technical math sections. I am closing this thread.
 

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