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

R2.3

Problem Statement

Find a general solution. Check your answer by substitution.

a) $y^{″}+6y^{\prime }+8.96y=0$ (3-1)

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

Solution

The quadratic formula is necessary for these solutions:

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

Part a

Plugging into the quadratic formula:

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

This shows us that the roots of the equation are:

${\lambda }_1=-2.8,{\lambda }_2=-3.2$

Therefore, the general equation is:

$y=c_1{e}^{-2.8x}+c_2{e}^{-3.2x}$     (3-3)

Substitution

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

$y^{\prime }=-3.2c_1{e}^{-3.2x}-2.8c_2{e}^{-2.8x}$

$y^{″}=10.24c_1{e}^{-3.2x}+7.84c_2{e}^{-2.8x}$

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

$(10.24c_1{e}^{-3.2x}+7.84c_2{e}^{-2.8x})+6(-3.2c_1{e}^{-3.2x}-2.8c_2{e}^{-2.8x})+8.96(c_1{e}^{-3.2x}+c_2{e}^{-2.8x})=0$ (3-4)

Continuing to solve:

$19.2c_1{e}^{-3.2x}+16.8c_2{e}^{-2.8x}-19.2c_1{e}^{-3.2x}-16.8c_2{e}^{-2.8x}=0$ (3-5)

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

Part b

Plugging into the quadratic formula:

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

The roots are, therefore:

${\lambda }_1=-2-\pi i,{\lambda }_2=-2+\pi i$

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

$y=c_1\cos (\pi x)e^{-2x}+c_2\sin (\pi x)e^{-2x}$ (3-6)

Substitution

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

$y^{\prime }=-2(c_1\cos (\pi x)+c_2\sin (\pi x))e^{-2x}+(-c_1\pi \sin (\pi x)+c_2\pi \cos (\pi x))e^{-2x}$

$y^{″}=-2(-c_1\pi \sin (\pi x)+c_2\pi \cos (\pi x))e^{-2x}+(-c_1{\pi }^2\cos (\pi x)-c_2{\pi }^2\sin (\pi x))e^{-2x}$

Plugging into equation (3-2):

$-2(-c_1\pi \sin (\pi x)+c_2\pi \cos (\pi x))e^{-2x}+(-c_1{\pi }^2\cos (\pi x)-c_2{\pi }^2\sin (\pi x))e^{-2x}...$

$+4[-2(c_1\cos (\pi x)+c_2\sin (\pi x))e^{-2x}+(-c_1\pi \sin (\pi x)+c_2\pi \cos (\pi x))e^{-2x}]+({\pi }^2+4)(c_1\cos (\pi x)e^{-2x}+c_2\sin (\pi x)e^{-2x})=0$

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

$4(c_1\cos (\pi x)+c_2\sin (\pi x))e^{-2x}-2(-c_1\pi \sin (\pi x)+c_2\pi \cos (\pi x))e^{-2x}-2(-c_1\pi \sin (\pi x)+c_2\pi \cos (\pi x))e^{-2x}...$

$+(-c_1{\pi }^2\cos (\pi x)-c_2{\pi }^2\sin (\pi x))e^{-2x}+4-2(c_1\cos (\pi x)+c_2\sin (\pi x))e^{-2x})+...$

$(-c_1\pi \sin (\pi x)+c_2\pi \cos (\pi x))e^{-2x}+({\pi }^2+4)((c_1\cos (\pi x)+c_2\sin (\pi x))e^{-2x})=0$

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