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

R2.3

Problem Statement

Find a general solution. Check your answer by substitution.

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

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

Solution

The quadratic formula is necessary for these solutions:

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

Part a

Plugging into the quadratic formula:

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

This shows us that the roots of the equation are:

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

Therefore, the general equation is:

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

Substitution

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

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

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

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

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

Continuing to solve:

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

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

Part b

Plugging into the quadratic formula:

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

The roots are, therefore:

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

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

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

Substitution

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

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

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

Plugging into equation (3-2):

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

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

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

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

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

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

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

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