Repeated roots, non homogeneous - second order, reduction of order method

In summary: Thank you,one more thing: what do you do with the constant of integration when forming the general solution?
  • #1
shorty1
16
0
I semi understand the reduction of order method, and i understand the general solution for a 2nd order with repeated roots. however, i can't seem to form up the correct thing to solve this question, and research again proves futile. Any assistance will be appreciated.

Use the method of reduction of order to solve

y'' - 4y' + 4y = ex


when i do the auxiliary i get my roots to be -2, repeated. but from there i am not sure how to go on. i tried letting y1 = e2x and letting y = y1 v(x), and found y' and y'' to substitute back in the original equation to equate the coefficients, but that didn't work. I am now confused.

thanks for your assistance
 
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  • #2
shorty said:
I semi understand the reduction of order method, and i understand the general solution for a 2nd order with repeated roots. however, i can't seem to form up the correct thing to solve this question, and research again proves futile. Any assistance will be appreciated.

Use the method of reduction of order to solve

y'' - 4y' + 4y = ex


when i do the auxiliary i get my roots to be -2, repeated. but from there i am not sure how to go on. i tried letting y1 = e2x and letting y = y1 v(x), and found y' and y'' to substitute back in the original equation to equate the coefficients, but that didn't work. I am now confused.

thanks for your assistance

Let's start from the homogeneous DE...

$\displaystyle y^{\ ''} - 4 y^{\ '} +4\ y=0$ (1)

You have found that $\displaystyle u(x)=e^{2 x}$ is a solution of (1) and that is correct. Now a general procedure to find another solution $v(x)$ of (1) independent from $u(x)$ is illustrated in...

http://www.mathhelpboards.com/showthread.php?605-Real-double-roots-question&p=3605#post3605

Kind regards

$\chi$ $\sigma$
 
  • #3
shorty said:
I semi understand the reduction of order method, and i understand the general solution for a 2nd order with repeated roots. however, i can't seem to form up the correct thing to solve this question, and research again proves futile. Any assistance will be appreciated.

Use the method of reduction of order to solve

y'' - 4y' + 4y = ex


when i do the auxiliary i get my roots to be -2, repeated. but from there i am not sure how to go on. i tried letting y1 = e2x and letting y = y1 v(x), and found y' and y'' to substitute back in the original equation to equate the coefficients, but that didn't work. I am now confused.

thanks for your assistance

1. The roots are +2.

2.

\(y=e^{2x}v(x)\)

\(y'=2e^{2x}v(x)+e^{2x}v'(x)\)

\(y''=4e^{2x}v(x)+4e^{2x}v'(x)+e^{2x}v''(x)\)

Now putting these into the equation we get:

\( [4e^{2x}v(x)+4e^{2x}v'(x)+e^{2x}v''(x)] -4[2e^{2x}v(x)+e^{2x}v'(x)] +4[e^{2x}v(x)]=e^{x}\)

which reduces to:

\(v''(x)=e^{-x}\)


CB
 
  • #4
Thank you,

one more thing: what do you do with the constant of integration when forming the general solution?

I have $$ y = C_1 e^{2x} c_2 e^{-x} $$ as my general solution. what should i have done with the Constant of integration?
CaptainBlack said:
1. The roots are +2.

2.

\(y=e^{2x}v(x)\)

\(y'=2e^{2x}v(x)+e^{2x}v'(x)\)

\(y''=4e^{2x}v(x)+4e^{2x}v'(x)+e^{2x}v''(x)\)

Now putting these into the equation we get:

\( [4e^{2x}v(x)+4e^{2x}v'(x)+e^{2x}v''(x)] -4[2e^{2x}v(x)+e^{2x}v'(x)] +4[e^{2x}v(x)]=e^{x}\)

which reduces to:

\(v''(x)=e^{-x}\)


CB
 
  • #5
shorty said:
Thank you,

one more thing: what do you do with the constant of integration when forming the general solution?

I have $$ y = C_1 e^{2x} c_2 e^{-x} $$ as my general solution. what should i have done with the Constant of integration?

From \(v''(x)=e^{-x} \) you get \(v(x)=e^{-x}+(ax+b)\) which when recombined with \(e^{2x}\) gives a general solution: \[y(x)=e^{2x}[e^{-x}+ax+b)]=e^x+axe^{2x}+be^{2x}\]

The first term on the right is a particular integral and the remaining two terms are the general solution to the homogeneous equation.

CB
 

Related to Repeated roots, non homogeneous - second order, reduction of order method

1. What are repeated roots in the context of second order non-homogeneous functions?

Repeated roots refer to the situation where the characteristic equation of a second order non-homogeneous function has a repeated root. This means that the roots of the characteristic equation are the same, resulting in a repeated term in the general solution of the function.

2. How do repeated roots affect the solution of a second order non-homogeneous function?

Repeated roots can affect the solution of a second order non-homogeneous function by making the general solution more complicated. Instead of having distinct roots, the repeated root introduces a repeated term in the general solution, which can make it more difficult to find the particular solution for a specific initial condition.

3. What is the reduction of order method for solving second order non-homogeneous functions?

The reduction of order method is a technique used to simplify the solution of a second order non-homogeneous function with repeated roots. It involves finding a new function that satisfies the same differential equation as the original function, but with one less unknown constant. This simplifies the process of finding the particular solution for a given initial condition.

4. How does the reduction of order method work for solving second order non-homogeneous functions with repeated roots?

The reduction of order method works by substituting the repeated root into the general solution of the original function, which results in a simpler solution with one less unknown constant. This new function can then be used to find the particular solution for a given initial condition, making the overall solution process more efficient.

5. When should the reduction of order method be used to solve second order non-homogeneous functions with repeated roots?

The reduction of order method should be used when solving second order non-homogeneous functions with repeated roots in order to simplify the solution process. This method is especially useful when the repeated root results in a complex general solution, as it allows for a more efficient and straightforward way to find the particular solution for a specific initial condition.

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