User:Egm4313.s12.team8.koester/R1.5.2

Problem 5
Problem found on page 59, problem number 5

$$ \displaystyle 5. \; {y}''+2\pi {y}'+\pi ^{2}y=0 $$

The characteristic equation that corresponds to this kind of problem is:

$$ \displaystyle \; {y}''+a {y}'+b y=0 $$

So for this problem:

$$ \displaystyle \;  a=2\pi  $$  and  $$ \displaystyle  \;   b=\pi ^{2} $$

Substiting into the equation we have a term $$ \displaystyle a^{2}-4b $$ that tells us the case number we are dealing with

$$ \displaystyle \;  a^{2}-4b= (2\pi)^{2}-4(\pi ^{2})= 4\pi ^{2}- 4\pi ^{2} =0 $$

Since this term equals 0, we have Case II and therefore a real double root.

Therefore:

$$ \displaystyle \;  \lambda_1= -a/2= - 2\pi/2 = -\pi $$

The first solution is:

$$ \displaystyle \;  y_1= e ^{\lambda_1x}= e ^{-\pi x} $$

You can obtain the second solution by using the method of reduction of order

$$ \displaystyle \; y_2= uy_1  $$

$$ \displaystyle \; {y_2}'= {u}'y_1 + u{y_1}'  $$

$$ \displaystyle \; {y_2}= {u}y_1 +2{u}'{y_1}'+ u{y_1}''  $$

Substituting these derivatives into the original characteristic equation we get:

$$ \displaystyle \; ({u}y_1 +2{u}'{y_1}'+ u{y_1}) + a({u}'y_1 + u{y_1}') + buy_1 =0  $$

Collecting $$ \displaystyle u, {u}', {u}''  $$ terms together you obtain:

$$ \displaystyle \; {u}y_1 + {u}'(2{y_1}'+ay_1) + u({y_1}+a{y_1}'+by_1)=0  $$

The second and third terms will disappear to leave you with:

$$ \displaystyle \; {u}y_1=0 $$ and therefore $$ \displaystyle  \; {u}=0  $$

Integrate twice to get the expressions:

$$ \displaystyle \; {u}'=c_1  $$

$$ \displaystyle \; u=c_1x+c_2  $$

We can now simply choose for $$ \displaystyle \; c_1=1 $$ and $$ \displaystyle  \; c_2=0 $$

So now:

$$ \displaystyle \; u=x  $$ and therefore $$ \displaystyle  \;y_2=xy_1   $$

Therefore:

$$ \displaystyle \; y_2= xe^ {-\pi x}  $$

And the General Solution becomes:

$$ \displaystyle \; y=(c_1+c_2x)e^ {-\pi x}   $$

Solved and typed by: Egm4313.s12.team8.koester 02:32, 31 January 2012 (UTC)

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