Find a linear homogeneous equation with given general solution

[pedro123]

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

 

[pedro123]

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.

 

[pedro123]

is the the differential equation -y''''-y''=0 I am not sure

 

[jeannea]

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

 

[pedro123]

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.

 

[jeannea]

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

 

[pedro123]

making the differential equation

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

is this alright and sorry about my mistake in factoring

 

[jeannea]

The differential equation would be

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

Remember that r=r^{1} and 1=r^{0}

 

[pedro123]

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

 

[pedro123]

and thanks for helping me so far

 

[pedro123]

I know that 2 is a constant but can my particular solution be

Yh=Acos4t +Bsin4t

 

[Mark44]

Problems such as these should be posted in the Homework & Coursework section, not in the technical math sections. I am closing this thread.