Homework Week 3 Solutions

P144: 1, 10, 15, 17; P145: 27, 28.
P155: 6, 8, 13, 21; P156: 33.
P163: 7, 10, 20; P171: 12.

Posted 20110226.

page 144 number 1
$r^{2}+2r-3=0$

$(r-1)(r+3)=0$

$y=c_{1}e^{t}+c_{2}e^{-3t}$

page 144 number 10

$\left(r+1\right)\left(r+3\right)=0$

$y=c_{1}e^{-t}+c_{2}e^{-3t}$

$1=y(0)=c_{1}+c_{2}$

$1=y^{\prime}(0)=-c_{1}-3c_{2}$

$2=-2c_{2}$

$c_{2}=-1$

$c_{1}=2$

$y=2e^{-t}-e^{-3t}$

page 144 number 15

$(r-1)(r+9)=0$

$y=c_{1}e^{t}+c_{2}e^{-9t}$

$y^{\prime}=c_{1}e^{t}-9c_{2}e^{-9t}$

$1=y(1)=c_{1}e+c_{2}e^{-9}$

$0=y^{\prime}(1)=c_{1}e-9c_{2}e^{-9}$

$1=10c_{2}e^{-9}$

$c_{2}=\frac{e^{9}}{10}$

$c_{1}=\frac{9}{10}e^{-1}$

$y=\frac{9}{10}e^{t-1}+\frac{1}{10}e^{9-9t}$

page 144 number 17

$(r-2)(r+3)=0$

$r^{2}+r-6=0$

$y^{\prime\prime}+y^{\prime}-6y=0$

page 145 number 27

$y=C$ a constant

$y^{\prime}=y^{\prime\prime}=0$

Then

$c\cdot C=d$

If $c\neq0$ then $C=\frac{d}{c}$

If $c=0$ then there is only a solution when $d=0$ and in that case
any constant is a solution.

For part b we have

$ay^{\prime\prime}+by^{\prime}+cy=d$ and $ay_{e}^{\prime\prime}+by_{e}^{\prime}+cy_{e}=d$

So then

$a(y-y_{e})^{\prime\prime}+b(y-y_{e})^{\prime}+c(y-y_{e})=0$

Hence

$aY^{\prime\prime}+bY^{\prime}+cY=0$

page 145 number 28

One important part is that $a>0$. So just divide by $a$ to get $y^{\prime\prime}+By^{\prime}+Cy=0$

In all situations, we need two different real roots so we know that
$B^{2}-4C>0$. So then $B^{2}>4C$.

Part a)

$(r-n_{1})(r-n_{2})=r^{2}-(n_{1}+n_{2})r+n_{1}n_{2}$

Thus $B>0$ and $C>0$. Thus we require at least that $B^{2}>4C>0$.

Maybe this is enough. We check that $\frac{-B+\sqrt{B^{2}-4C}}{2}<0$ This is equivalent to checking that $0<\sqrt{B^{2}-4C}<B$ So this is the same as showing that $B^{2}-4C < B^{2}$ And this is true because $0<4C$. Thus we conclude $B>0$, $C>0$, and $B^{2}>4C$.

Part b)

$(r-n)(r-p)=r^{2}-(n+p)r+np$

So $B$ can be positive or negative, depending on which root is bigger.

But $C<0$. $-B-\sqrt{B^{2}-4C}<0$. And $-B+\sqrt{B^{2}-4C}>0$

So $-\sqrt{B^{2}-4C} < B<\sqrt{B^{2}-4C}$ Thus $|B|<\sqrt{B^{2}-4C}$ Which is the same as asking for $B^{2} < B^{2}-4C$ Which comes from $4C<0$. Thus $C<0$ is a sufficient and necessary condition. For this situation we require $C<0$. Part c) $(r-p_{1})(r-p_{2})=r^{2}-(p_{1}+p_{2})r+p_{1}p_{2}$ $C>0$. $B<0$. $B^{2}>4C$.

Thus

$4C>0$

Then $0>-4C$

Then $B^{2}>B^{2}-4C$. Then $|B|>\sqrt{B^{2}-4C}$. So $B>\sqrt{B^{2}-4C}$
or $-B<-\sqrt{B^{2}-4C}$. Then $-B<-\sqrt{B^{2}-4C}$ and $-B<\sqrt{B^{2}-4C}$. We conclude that $-B\pm\sqrt{B^{2}-4C}<0$. Thus $C>0$, $B<0$, and $B^{2}-4C>0$ is a necessary and sufficient
condition for two real different positive roots.

page 155 number 6
$\left|\begin{array}{cc}
\cos^{2}\theta & 1+\cos2\theta\\
-2\cos\theta\sin\theta & -2\sin2\theta\end{array}\right|=\left|\begin{array}{cc}
\frac{1+\cos2\theta}{2} & 1+\cos2\theta\\
-\sin2\theta & -2\sin2\theta\end{array}\right|=0$

Update to Problem 6 [20110302]:
There are some alternate ways to go about the problem.
The two most useful trigonometric identities to use in the problem are
1) $\cos^2\theta=\frac{1+\cos2\theta}{2}$
2) $\sin2\theta=2\sin\theta\cos\theta$
Alternatively to 1) is
1') $2\cos^2\theta=1+\cos2\theta$
Other identities may be used, but tend to make the problem longer.

page 155 number 21

One way to do the problem is just to straight up compute the Wronskian:

$W(y_{3},y_{4})=\left|\begin{array}{cc}
a_{1}y_{1}+a_{2}y_{2} & b_{1}y_{1}+b_{2}y_{2}\\
a_{1}y_{1}^{\prime}+a_{2}y_{2}^{\prime} & b_{1}y_{1}^{\prime}+b_{2}y_{2}^{\prime}\end{array}\right|=a_{1}y_{1}b_{1}y_{1}^{\prime}+a_{2}y_{2}b_{1}y_{1}^{\prime}+a_{1}y_{1}b_{1}y_{1}^{\prime}+\dots$

Much easier is to use linear algebra.

$W(y_{3},y_{4})=\left|\begin{array}{cc}
a_{1}y_{1}+a_{2}y_{2} & b_{1}y_{1}+b_{2}y_{2}\\
a_{1}y_{1}^{\prime}+a_{2}y_{2}^{\prime} & b_{1}y_{1}^{\prime}+b_{2}y_{2}^{\prime}\end{array}\right|=\left|\begin{array}{cc}
y_{1} & y_{2}\\
y_{1}^{\prime} & y_{2}^{\prime}\end{array}\right|\left|\begin{array}{cc}
a_{1} & b_{1}\\
a_{2} & b_{2}\end{array}\right|=W(y_{1},y_{2})\left(a_{1}b_{2}-a_{2}b_{1}\right)$

Whether or not $y_{3}$ and $y_{4}$ are also a fundamental set of
solutions depends on whether or not $a_{1}b_{2}-a_{2}b_{1}=0$.

page 156 number 33

$[p(t)y^{\prime}]^{\prime}+q(t)y=0$

$p$ is differentialable so we get $p(t)y^{\prime\prime}+p^{\prime}(t)y^{\prime}+q(t)y=0$

$p>0$ so we can divide by it to get $y^{\prime\prime}+\frac{p^{\prime}(t)}{p(t)}y^{\prime}+\frac{q(t)}{p(t)}y=0$

Use Abel's theorem.

$W(t)=e^{-\int\frac{p^{\prime}(t)}{p(t)}dt}$

Use u-substitution.

$W(t)=ce^{-\int\frac{1}{p(t)}dp(t)}=ce^{-\ln p(t)}=ce^{\ln p(t)^{-1}}=cp(t)^{-1}=c/p(t)$.

Note: Typically $f(t)^{-1}=\frac{1}{f(t)}.$ And $f^{-1}(t)$ is the
inverse of $f$. If you are ever unsure about notation, ask.

page 163 number 7

$r=\frac{2\pm\sqrt{4-8}}{2}=\frac{2\pm2i}{2}=1\pm i$

$y=c_{1}e^{t}\cos t+c_{2}e^{t}\sin t$

page 163 number 10

$r=\frac{-2\pm\sqrt{4-8}}{2}=-1\pm i$

$y=c_{1}e^{-t}\cos t+c_{2}e^{-t}\sin t$

page 163 number 20

$r=\frac{\pm\sqrt{-4}}{2}=\pm i$

$y=c_{1}\cos t+c_{2}\sin t$

$2=y(\pi/3)=c_{1}\cdot\frac{1}{2}+c_{2}\cdot\frac{\sqrt{3}}{2}$

$y^{\prime}=-c_{1}\sin t+c_{2}\cos t$

$-4=y^{\prime}(\pi/3)=-c_{1}\cdot\frac{\sqrt{3}}{2}+c_{2}\cdot\frac{1}{2}$

$2\sqrt{3}-4=c_{2}\left(\frac{3}{2}+\frac{1}{2}\right)$

$c_{2}=\sqrt{3}-2$

$2=c_{1}\cdot\frac{1}{2}+\frac{3-2\sqrt{3}}{2}$

$4=c_{1}+(3-2\sqrt{3})$

$c_{1}=1+2\sqrt{3}$

$y=(1+2\sqrt{3})\cos t+(\sqrt{3}-2)\sin t$

page 171 number 12
Corrected solution to number 12 [20110302]
$r=3$

$y=c_{1}e^{3t}+c_{2}te^{3t}$

$0=y(0)=c_{1}$

$y^{\prime}=3c_{1}e^{3t}+c_{2}e^{3t}+3c_{2}te^{3t}$

$2=y^{\prime}(0)=3c_{1}+c_{2}$

So $c_{2}=2$

$y=2te^{3t}$