MATH 308 Lecture 11

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Lecture Notes

Announcements:

• Quiz this Friday over section 3.1 and 3.3
• Homework is due Monday

Homogeneous DEs with Constant Coefficients

$a{y}^{″}+b{y}^{\prime }+cy=0$

Characteristic equation is

$a{r}^2+br+c=0$

General solutions are of the form

$y=c_1{\mathrm{e}}^{r_{1}t}+c_2{\mathrm{e}}^{r_{2}t}$

If $r$ values are complex conjugates, then real-valued solution is of form

$y=c_1\cos \theta +c_2\sin \theta$

where the cosine and sine come from Euler's formula:

${\mathrm{e}}^{it}=\cos t+i\sin t$

Exercise 5

$\begin{array}{rl}{y}^{″}-2y^{\prime }+5y& =0\\ r^2-2r+5& =0\\ r& =1\pm 2i\\ y_1& ={\mathrm{e}}^{t+2it}\\ & ={\mathrm{e}}^t(\cos (2t)+i\sin (2t))\\ y_2& ={\mathrm{e}}^{t-2it}\\ & ={\mathrm{e}}^t(\cos (-2t)+i\sin (-2t))\\ \frac{y_1+y_2}{2}& ={\mathrm{e}}^t\cos 2t\end{array}$

What if the equation has two identical roots?

In general, if $r=a\pm bi$ are roots of the characteristic equation, then the general solution is $y=c_1{\mathrm{e}}^{at}\cos (bt)+c_2{\mathrm{e}}^{at}\sin (bt)$

Section 3.4

$y^{″}-4y^{\prime }+4y=0$

Characteristic equation $r^2-4r+4=(r-2{)}^2=0$ has two roots at $r=2$.

Let $y_1={\mathrm{e}}^{2t}$ and $y_2=t{\mathrm{e}}^{2t}$

The general solution is $y=c_1{\mathrm{e}}^{2t}+c_{2}t{\mathrm{e}}^{2t}$

Exercise 7a

Find the solution to the initial value problem $y^{″}+y^{\prime }-2y=0$, where $y(0)=1$ and $y^{\prime }(0)=3$.

$\begin{array}{rl}{r}^2+r-2& =0\\ r& \in \{-2,1\}\end{array}$

General solution is $y=c_1{\mathrm{e}}^t+c_2{\mathrm{e}}^{-2t}$. To find particular solution, we need $y^{\prime }=c_1{\mathrm{e}}^t-2c_2{\mathrm{e}}^{-2t}$

$\begin{array}{rl}1& =c_1+c_2\\ 3& =c_1-2c_2\\ c_1& =\frac{5}{3}\\ c_2=-\frac{2}{3}\end{array}$

So the particular solution is $y=\frac{5}{3}{\mathrm{e}}^t-\frac{2}{3}{\mathrm{e}}^{-2t}$

Exercise 7b

Find the solution to the initial value problem $9y^{″}-12y^{\prime }+4y=0$, where $y(0)=2$ and $y^{\prime }(0)=-1$.

$\begin{array}{rl}9r^2-12r+4& =0\\ (3r-2{)}^2& =0\\ r& \in \{\frac{2}{3},\frac{2}{3}\}\end{array}$

The general solution is $y=c_1{\mathrm{e}}^{\frac{2t}{3}}+c_{2}t{\mathrm{e}}^{\frac{2t}{3}}$

We can already find that $y(0)=c_1=2$, so ...

So $y=2{\mathrm{e}}^{\frac{2t}{3}}-\frac{7}{3}t{\mathrm{e}}^{\frac{2t}{3}}$

Exercise 8

$y^{″}+2y^{\prime }+2y=0$, $y(\frac{\pi }{4})=2$ and $y^{\prime }(\frac{\pi }{4}=-2$.

$\begin{array}{rl}{r}^2+2r+2& =0\\ r& =-1\pm i\end{array}$

General solution is $y=c_1{\mathrm{e}}^{-t}\cos t+c_2{\mathrm{e}}^{-t}\sin t$.

Find particular solution by plugging in initial conditions:

$\begin{array}{rl}y\left(\frac{\pi }{4}\right)=2& =c_1{\mathrm{e}}^{-\frac{\pi }{4}}\frac{\sqrt{2}}{2}+c_2{\mathrm{e}}^{-\frac{\pi }{4}}\frac{\sqrt{2}}{2}\\ c_1+c_2=2\sqrt{2}{\mathrm{e}}^{\frac{\pi }{4}}\end{array}$

Differentiate the general solution: $y^{\prime }=-c_1{\mathrm{e}}^{-t}\cos t-c_1{\mathrm{e}}^{-t}\sin t-c_2{\mathrm{e}}^{-t}\sin t+c_2{\mathrm{e}}^{-t}\cos t$

And plug in initial condition.

$y^{\prime }\left(\frac{\pi }{2}\right)=-2c_1{\mathrm{e}}^{-\frac{\pi }{4}}\frac{\sqrt{2}}{2}$

Now we can solve for $c_1=\sqrt{2}{\mathrm{e}}^{\frac{\pi }{4}}$ and $c_2=\sqrt{2}{\mathrm{e}}^{\frac{\pi }{4}}$, so the particular solution is

$y=\sqrt{2}{\mathrm{e}}^{\frac{\pi }{4}}{\mathrm{e}}^{-t}\cos t+\sqrt{2}{\mathrm{e}}^{\frac{\pi }{4}}{\mathrm{e}}^{-t}\sin t$

Theorem of Existence and Uniqueness

Consider the initial value problem

$y^{″}+p(t)y^{\prime }+q(t)y=g(t)y(t_0)=y_0{y}^{\prime }(t_0)=y{\text{'}}_0$

Where $p$, $q$, and $g$ are continuous on an open interval $I$ that contains the point $t_0$. Then there exists exactly one solution $y=\mathrm{\Phi }(t)$ of this problem, and the solution exists throughout the interval $I$.