Find a linear homogeneous equation with given general solution

  • Thread starter pedro123
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  • #1
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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
 

Answers and Replies

  • #2
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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.
 
  • #3
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is the the differential equation -y''''-y''=0 im not sure
 
  • #4
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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
 
  • #5
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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.
 
  • #6
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Not quite. If you multiply (r-1)(r+1)[itex]^{3}[/itex], you should get
(r-1)(r[itex]^{3}[/itex]+3r[itex]^{2}[/itex]+3r+1)=r[itex]^{4}[/itex]+2r[itex]^{3}[/itex]-2r-1
 
  • #7
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making the differential equation

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

is this alright and sorry about my mistake in factoring
 
  • #8
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The differential equation would be

y[itex]^{(4)}[/itex]+2y[itex]^{(3)}[/itex]-2y'-y

Remember that r=r[itex]^{1}[/itex] and 1=r[itex]^{0}[/itex]
 
  • #9
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thanks can help solve this problem


I 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
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and thanks for helping me so far
 
  • #11
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I know that 2 is a constant but can my particular solution be

Yh=Acos4t +Bsin4t
 
  • #12
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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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