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

$a\,y''+b\,y'+c\,y=0$

Characteristic equation is

$a\,r^{2}+b\,r+c=0$

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:

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

Exercise 5

${\begin{aligned}y''-2y'+5y&=0\\r^{2}-2r+5&=0\\r&=1\pm 2i\\y_{1}&=\mathrm {e} ^{t+2i\,t}\\&=\mathrm {e} ^{t}\left(\cos {(2t)}+i\,\sin {(2t)}\right)\\y_{2}&=\mathrm {e} ^{t-2i\,t}\\&=\mathrm {e} ^{t}\left(\cos {(-2t)}+i\,\sin {(-2t)}\right)\\{\frac {y_{1}+y_{2}}{2}}&=\mathrm {e} ^{t}\cos {2t}\end{aligned}}$

What if the equation has two identical roots?

In general, if $r=a\pm b\,i$ are roots of the characteristic equation, then the general solution is $y=c_{1}\,\mathrm {e} ^{a\,t}\,\cos {(b\,t)}+c_{2}\,\mathrm {e} ^{a\,t}\,\sin {(b\,t)}$

Section 3.4

$y''-4y'+4y=0$

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 $y''+y'-2y=0$, where $y(0)=1$ and $y'(0)=3$.

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

General solution is $y=c_{1}\,\mathrm {e} ^{t}+c_{2}\,\mathrm {e} ^{-2t}$. To find particular solution, we need $y'=c_{1}\,\mathrm {e} ^{t}-2c_{2}\,\mathrm {e} ^{-2t}$

${\begin{aligned}1&=c_{1}+c_{2}\\3&=c_{1}-2c_{2}\\c_{1}&={\frac {5}{3}}\\c_{2}=-{\frac {2}{3}}\end{aligned}}$

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 $9y''-12y'+4y=0$, where $y(0)=2$ and $y'(0)=-1$.

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

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

$y''+2y'+2y=0$, $y({\tfrac {\pi }{4}})=2$ and $y'({\tfrac {\pi }{4}}=-2$.

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

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:

${\begin{aligned}y\left({\frac {\pi }{4}}\right)=2&=c_{1}\,\mathrm {e} ^{-{\frac {\pi }{4}}}\,{\frac {\sqrt {2}}{2}}+c_{2}\,\mathrm {e} ^{-{\frac {\pi }{4}}}\,{\frac {\sqrt {2}}{2}}\\c_{1}+c_{2}=2{\sqrt {2}}\,\mathrm {e} ^{\frac {\pi }{4}}\end{aligned}}$

Differentiate the general solution: $y'=-c_{1}\,\mathrm {e} ^{-t}\,\cos {t}-c_{1}\,\mathrm {e} ^{-t}\,\sin {t}-c_{2}\,\mathrm {e} ^{-t}\,\sin {t}+c_{2}\,\mathrm {e} ^{-t}\,\cos {t}$

And plug in initial condition.

$y'\left({\frac {\pi }{2}}\right)=-2c_{1}\,\mathrm {e} ^{-{\frac {\pi }{4}}}\,{\frac {\sqrt {2}}{2}}$

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

$y''+p(t)\,y'+q(t)\,y=g(t)\quad y(t_{0})=y_{0}\quad y'(t_{0})=y'_{0}$

Where $p$, $q$, and $g$ are continuous on an open interval $I$ that contains the point $t_{0}$. Then there exists exactly one solution $y=\Phi (t)$ of this problem, and the solution exists throughout the interval $I$.