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

shorty

New member
Feb 5, 2012
16
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
 

chisigma

Well-known member
Feb 13, 2012
1,704
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

$\chi$ $\sigma$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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
 

shorty

New member
Feb 5, 2012
16
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?
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
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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