Find the general solution

In summary, the conversation discusses finding the general solution for the differential equation y'' + 4y' +4y = 5xe^(-2x) using the method of undetermined coefficients. One solution is found to be (5/2)x^3*e^(-2x), while another source suggests the solution to be (5/6)x^3*e^(-2x). The conversation also briefly mentions finding the roots of the characteristic equation for the homogeneous part of the equation and discusses the process of finding the particular solution. Finally, there is a clarification on the constant term in the particular solution.
  • #1
Squeezebox
57
0

Homework Statement



Find the general solution

y'' + 4y' +4y = 5xe^(-2x)



The Attempt at a Solution



I got (5/2)x^3*e^(-2x) as a particular solution. But I checked online at wolfram alpha and it says the particular solution is (5/6)x^3*e^(-2x). Using method of undetermined coefficients.
 
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  • #2
Solve the homogeneous part of y'' + 4y' +4y = 0. What are the roots of the characteristic equation?
 
  • #3
rock.freak667 said:
Solve the homogeneous part of y'' + 4y' +4y = 0. What are the roots of the characteristic equation?

The homogeneous part gave me

y=c1e-2x+c2xe-2x

from the root (D+2)2.

Multiplying by another (D+2) annihilated the 5xe-2x, resulting in a new solution

y=c1e-2x+c2xe-2x+c3x2e-2x

is that right?
 
  • #4
Squeezebox said:
Multiplying by another (D+2) annihilated the 5xe-2x, resulting in a new solution

y=c1e-2x+c2xe-2x+c3x2e-2x


Your right side is xe-2x, since -2 is seen twice in the homogeneous part, you must multiply it by x2.


Can you show your work on how you got the constant to be 5/2?
 
  • #5
rock.freak667 said:
Your right side is xe-2x, since -2 is seen twice in the homogeneous part, you must multiply it by x2.


Can you show your work on how you got the constant to be 5/2?

(D2+4D+4)(c3x2e-2x) = 5xe-2x

c3(4x2e-2x-8xe-2x+2e-2x+4(2xe-2x-2x2e-2x)+4x2e-2x) = 5xe-2x


Every thing but c3(2e-2x) reduces to zero

c3(2e-2x)=5xe-2x
c3=(5/2)x
 
  • #6
Squeezebox said:
(D2+4D+4)(c3x2e-2x) = 5xe-2x

c3(4x2e-2x-8xe-2x+2e-2x+4(2xe-2x-2x2e-2x)+4x2e-2x) = 5xe-2x


Every thing but c3(2e-2x) reduces to zero

c3(2e-2x)=5xe-2x
c3=(5/2)x

I thought you got yp=c3x3e-2x
 
  • #7
rock.freak667 said:
I thought you got yp=c3x3e-2x

yp=c3x2e-2x
yp=(5/2)x*x2e-2x
yp=(5/2)x3e-2x
 

1. What does "find the general solution" mean?

"Find the general solution" refers to finding a mathematical expression or formula that satisfies a given differential equation or system of equations. This general solution can then be used to find specific solutions for different initial conditions.

2. How is the general solution different from a specific solution?

The general solution is a mathematical expression that satisfies a given differential equation or system of equations for any set of initial conditions. A specific solution, on the other hand, is a solution that satisfies the equation for a specific set of initial conditions.

3. Why is it important to find the general solution?

Finding the general solution allows us to solve a differential equation or system of equations for any set of initial conditions. This is important because real-world problems often have varying initial conditions, and the general solution provides a way to find specific solutions for these different scenarios.

4. What techniques are used to find the general solution?

There are various techniques that can be used to find the general solution, such as separation of variables, integrating factors, and the method of undetermined coefficients. The specific technique used will depend on the type of differential equation or system of equations being solved.

5. Can the general solution be found for all differential equations?

No, not all differential equations have a general solution. Some equations may have no solution, while others may have only specific solutions. Additionally, some equations may have a general solution that cannot be expressed in terms of elementary functions and require more advanced mathematical techniques to solve.

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