Differential Equations, Particular and Complimentary solutions

In summary, the conversation discusses solving a differential equation using a transformation method. The first step is to find the complementary solution by setting the given equation equal to zero and solving for the roots. Then, the transformation y(x)=(e^(-bx))*v(x) is introduced into the equation and the particular solution is found by solving for v(x). The process involves finding the derivatives of y(x) and substituting them into the equation. The final particular solution is y(x)=((4/3)x^1.5)*e^(-bx).
  • #1
JefeNorte
9
0

Homework Statement


Given the differential equation for y=y(x)
(1) L[y]=y"+2by'+yb^2=(e^(-bx))/(x^2) x>0
a)find the complementary solution of (1) by solving L[y]=0
b)Solve (1) by introducing the transformation y(x)=(e^(-bx))*v(x) into (1) and obtaining and solving completely a differential equation for v(x). Use this to identify the particular solution

The Attempt at a Solution



Follwing the steps outline at http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
I converted y"+by'+yb^2 to r^2+2br+b^2=0

By factoring I determined that r1=r2=-b
so the complimentary solution should be y=c1*e^(-bt)+c2*e^(-bt)

Is this the right way to solve for a complimentary solution? If so how do I "induce the transformation to solve fro the particular solution?"
 
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  • #2
No, that is not correct. the complimentary solution is y= [itex]C_1e^{-bt}+ C_2t e^{-bt}[/itex].

How about doing what the problem asks you to do: "Solve (1) by introducing the transformation y(x)=(e^(-bx))*v(x) into (1)"?

If y(x)= [itex]e^{-bx}v(x)[/itex], what are y' and y"? Put those into the equation and see what equation you get for v.

Don't forget that the equation is 2by', not the by' you wrote in "I converted---". With the "2", it reduces nicely. Without, it's a mess!
 
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  • #3
So after introducing y(x)= v(x)*e^(-bx) to the initial equation I came up with v"=1/(x^2)

Using this I came up with v(x)=(4/3)x^(1.5)

So y(x)=((4/3)x^1.5)*e^(-bx) is the particular solution? I apologize for being so dense but I don't actually take this class until next semester, I am just trying to get ahead. Thanks
 

Related to Differential Equations, Particular and Complimentary solutions

1. What is the difference between particular and complimentary solutions in differential equations?

The particular solution of a differential equation is a specific solution that satisfies both the differential equation and any initial conditions. It is obtained by substituting the initial conditions into the general solution. On the other hand, the complimentary solution is the general solution that satisfies the homogeneous form of the differential equation. It contains a constant of integration and can be used to generate the particular solution.

2. How do you find the particular solution of a differential equation?

To find the particular solution, you first need to solve the homogeneous form of the differential equation. Then, you can use the method of undetermined coefficients or variation of parameters to find a particular solution that satisfies both the differential equation and any given initial conditions.

3. What is the order of a differential equation?

The order of a differential equation is the highest derivative present in the equation. For example, a first-order differential equation contains only the first derivative, while a second-order differential equation contains both the first and second derivatives.

4. What is the difference between an ordinary and a partial differential equation?

An ordinary differential equation involves only one independent variable, while a partial differential equation involves more than one independent variable. This means that the solution of an ordinary differential equation is a function of a single variable, while the solution of a partial differential equation is a function of multiple variables.

5. Can all differential equations be solved analytically?

No, not all differential equations have analytical solutions. Some differential equations, especially those with complex or non-linear terms, can only be solved numerically using approximation methods. However, some types of differential equations, such as linear first-order equations, have well-known analytical solutions.

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