User:Egm4313.s12.team8.colocar/R2.5

Problem
Kreyszig 2011 p.59 pbs.16-17

FIND AN ODE $$ \displaystyle y'' + ay' + by = 0 $$    (1)   for the given basis.

16. $$ \displaystyle e^{2.6x}, e^{-4.3x} $$

17. $$ \displaystyle e^{- \sqrt{5}x}, xe^{- \sqrt{5}x} $$

Solution
 16.  

Note: Given $$ \displaystyle y = e^{\lambda x} $$ as a solution of the second-order homogeneous linear ODE, the characteristic equation (or auxiliary equation) is defined as

$$ \displaystyle \lambda^2 + a\lambda + b = 0. $$  [K2011 p54]

The problem gives us two distinct real-roots:

$$ \displaystyle \lambda_{1} = 2.6 $$ and $$ \displaystyle \lambda_{2} = -4.3. $$

Solving the quadratic equation above gives us $$ \displaystyle \lambda_{1} = \frac {-a + \sqrt{a^2 - 4b}} 2 = \frac {1}{2} (-a + \sqrt{a^2 - 4b}) = 2.6, $$ and $$ \displaystyle \lambda_{2} = \frac {-a - \sqrt{a^2 - 4b}} 2 = \frac {1}{2} (-a - \sqrt{a^2 - 4b}) = -4.3 $$

or $$ \displaystyle \lambda_{1} = (-a + \sqrt{a^2 - 4b}) = 5.2, $$ and $$ \displaystyle \lambda_{2} = (-a - \sqrt{a^2 - 4b}) = -8.6. $$

Summing $$ \displaystyle \lambda_{1} $$ and $$ \displaystyle \lambda_{2} $$ we can eliminate the discriminant $$ -a + \sqrt{a^2 - 4b} $$ and solve for $$ \displaystyle a $$ $$ \displaystyle -2a = -3.4 $$ $$ \displaystyle a = 1.7 $$

Plugging $$ \displaystyle a $$ back into either $$ \displaystyle \lambda_1 $$ or $$ \displaystyle \lambda_2, $$ we can solve for $$ \displaystyle b $$

$$ \displaystyle -1.7 + \sqrt{1.7^2-4b} = 5.2$$ $$ \displaystyle     \sqrt{2.89-4b} = 6.9$$ $$ \displaystyle 2.89 - 4b = 44.61$$ $$ \displaystyle -4b = 44.72$$ $$ \displaystyle b = -11.18$$

Now plugging $$ \displaystyle a $$ and $$ \displaystyle b $$ back into the general ODE, we have $$ \displaystyle y'' + 1.7y' - 11.18y = 0 $$

 17.  

Given the basis: $$ \displaystyle e^{- \sqrt{5}x}, xe^{- \sqrt{5}x}, $$ we know we get only one root $$ \displaystyle \lambda = \lambda_1 = \lambda_2 = - \frac {a}{2}. $$ [K2011 p.55]

Thus, the discriminant $$ \displaystyle a^2 - 4b = 0. $$    (2)  

Knowing $$ \displaystyle \lambda = - \frac {a}{2} $$ for the given basis, we can say $$ \displaystyle e^{- \frac {a}{2} x} = e^{- \sqrt{5}x} $$

to solve for unknown $$ \displaystyle a. $$

$$ \displaystyle - \frac {a}{2} = - \sqrt {5} $$ $$ \displaystyle a = 2 \sqrt {5} $$

Plugging $$ \displaystyle a $$ into  (2)  gives us $$ \displaystyle ({2 \sqrt {5}})^2 - 4b = 0 $$ $$ \displaystyle b = \frac {20}{4} = 5. $$

Now substituting $$ \displaystyle a $$ and $$ \displaystyle b $$ into  (1)  gives us our ODE: $$ \displaystyle y'' + 2 \sqrt{5} y' + 5y = 0 $$