# University of Florida/Egm4313/s12.team8.dupre/R2.3

## R2.3

### Problem Statement

a) ${\displaystyle \displaystyle y''+6y'+8.96y=0}$ (3-1)

b)${\displaystyle \displaystyle y''+4y'+(\pi ^{2}+4)y=0}$ (3-2)

### Solution

The quadratic formula is necessary for these solutions:

${\displaystyle \displaystyle {\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$

#### Part a

${\displaystyle \displaystyle {\frac {-6\pm {\sqrt {6^{2}-(4)(1)(8.96)}}}{2(1)}}={\frac {-6\pm .4}{2}}}$

This shows us that the roots of the equation are:

${\displaystyle \displaystyle \lambda _{1}=-2.8,\lambda _{2}=-3.2}$

Therefore, the general equation is:

${\displaystyle \displaystyle y=c_{1}e^{-2.8x}+c_{2}e^{-3.2x}}$     (3-3)

##### Substitution

We need to first find the first and second derivatives of equation (3-3):

${\displaystyle \displaystyle y'=-3.2c_{1}e^{-3.2x}-2.8c_{2}e^{-2.8x}}$

${\displaystyle \displaystyle y''=10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x}}$

Plugging into equation (3-1), we find:

${\displaystyle \displaystyle (10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x})+6(-3.2c_{1}e^{-3.2x}-2.8c_{2}e^{-2.8x})+8.96(c_{1}e^{-3.2x}+c_{2}e^{-2.8x})=0}$ (3-4)

Continuing to solve:

${\displaystyle \displaystyle 19.2c_{1}e^{-3.2x}+16.8c_{2}e^{-2.8x}-19.2c_{1}e^{-3.2x}-16.8c_{2}e^{-2.8x}=0}$ (3-5)

This shows that the general equation is correct, since everything cancels out to 0.

#### Part b

${\displaystyle \displaystyle {\frac {4\pm {\sqrt {4^{2}-4(1)(\pi ^{2}+4)}}}{2(1)}}={\frac {-4\pm (-4)(1))(pi^{2}+4)}{2(1)}}}$

The roots are, therefore:

${\displaystyle \displaystyle \lambda _{1}=-2-\pi i,\lambda _{2}=-2+\pi i}$

Therefore, the general solution to (3-2) is:

${\displaystyle \displaystyle y=c_{1}\cos(\pi x)e^{-2x}+c_{2}\sin(\pi x)e^{-2x}}$ (3-6)

##### Substitution

We must first find the first and second derivatives of equation (3-6):

${\displaystyle \displaystyle y'=-2(c_{1}\cos(\pi x)+c_{2}\sin(\pi x))e^{-2x}+(-c_{1}\pi \sin(\pi x)+c_{2}\pi \cos(\pi x))e^{-2x}}$

${\displaystyle \displaystyle y''=-2(-c_{1}\pi \sin(\pi x)+c_{2}\pi \cos(\pi x))e^{-2x}+(-c_{1}\pi ^{2}\cos(\pi x)-c_{2}\pi ^{2}\sin(\pi x))e^{-2x}}$

Plugging into equation (3-2):

${\displaystyle \displaystyle -2(-c_{1}\pi \sin(\pi x)+c_{2}\pi \cos(\pi x))e^{-2x}+(-c_{1}\pi ^{2}\cos(\pi x)-c_{2}\pi ^{2}\sin(\pi x))e^{-2x}...}$

${\displaystyle \displaystyle +4[-2(c_{1}\cos(\pi x)+c_{2}\sin(\pi x))e^{-2x}+(-c_{1}\pi \sin(\pi x)+c_{2}\pi \cos(\pi x))e^{-2x}]+(\pi ^{2}+4)(c_{1}\cos(\pi x)e^{-2x}+c_{2}\sin(\pi x)e^{-2x})=0}$

Finally, plugging (3-6) and it's first and second derivatives into equation (3-2), we find:

${\displaystyle \displaystyle 4(c_{1}\cos(\pi x)+c_{2}\sin(\pi x))e^{-2x}-2(-c_{1}\pi \sin(\pi x)+c_{2}\pi \cos(\pi x))e^{-2x}-2(-c_{1}\pi \sin(\pi x)+c_{2}\pi \cos(\pi x))e^{-2x}...}$

${\displaystyle \displaystyle +(-c_{1}\pi ^{2}\cos(\pi x)-c_{2}\pi ^{2}\sin(\pi x))e^{-2x}+4-2(c_{1}\cos(\pi x)+c_{2}\sin(\pi x))e^{-2x})+...}$

${\displaystyle \displaystyle (-c_{1}\pi \sin(\pi x)+c_{2}\pi \cos(\pi x))e^{-2x}+(\pi ^{2}+4)((c_{1}\cos(\pi x)+c_{2}\sin(\pi x))e^{-2x})=0}$

Since this equals 0, we know that the general equation (3-6) is correct.