# Find a linear homogeneous equation with given general solution

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

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 im 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 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

and thanks for helping me so far

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

Yh=Acos4t +Bsin4t

Mark44
Mentor
Problems such as these should be posted in the Homework & Coursework section, not in the technical math sections. I am closing this thread.