Find a linear homogeneous equation with given general solution

In summary, the conversation involves finding a linear homogenous constant-coefficient differential equation with a given general solution. The conversation also includes attempts at finding the differential equation and factoring the characteristic equation. The conversation then shifts to a different problem involving a second order differential equation with imaginary roots and finding a particular solution.
  • #1
pedro123
14
0
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
 
Physics news on Phys.org
  • #2
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
is the the differential equation -y''''-y''=0 I am not sure
 
  • #4
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
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
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
making the differential equation

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

is this alright and sorry about my mistake in factoring
 
  • #8
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
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.
 

1. What is a linear homogeneous equation?

A linear homogeneous equation is an algebraic equation that contains only linear terms and has a constant term of zero. In other words, all the variables are raised to the first power and there are no constant terms. An example of a linear homogeneous equation is 2x + 3y = 0.

2. What does it mean for a linear homogeneous equation to have a general solution?

A general solution for a linear homogeneous equation is a set of values that satisfy the equation and can be written in the form of a linear combination of constants and variables. In other words, it is a solution that can be used for any given set of initial conditions.

3. How do you find the general solution of a linear homogeneous equation?

To find the general solution of a linear homogeneous equation, you need to first solve the equation for all of its variables. Then, you can express the solution in terms of arbitrary constants, which can be varied to obtain different solutions for different initial conditions.

4. Can a linear homogeneous equation have more than one general solution?

Yes, a linear homogeneous equation can have an infinite number of general solutions. This is because the solution can be expressed in terms of arbitrary constants, and there are an infinite number of possible values for these constants.

5. How can you use the general solution to find a specific solution for a linear homogeneous equation?

To find a specific solution for a linear homogeneous equation, you need to substitute specific values for the arbitrary constants in the general solution. These specific values can be chosen based on the initial conditions given in the problem. Once the constants are substituted, you will have a unique solution for the equation.

Similar threads

Replies
10
Views
1K
  • Differential Equations
Replies
3
Views
1K
Replies
3
Views
720
  • Differential Equations
Replies
4
Views
2K
Replies
2
Views
2K
Replies
6
Views
1K
Replies
4
Views
1K
  • Differential Equations
Replies
2
Views
3K
  • Differential Equations
Replies
2
Views
1K
  • Differential Equations
Replies
7
Views
308
Back
Top