# 7.1 Second-order linear equations  (Page 6/15)

 Page 6 / 15
$y\left(x\right)={c}_{1}{e}^{\left(\alpha +\beta i\right)x}+{c}_{2}{e}^{\left(\alpha -\beta i\right)x}.$

Using some smart choices for ${c}_{1}$ and ${c}_{2},$ and a little bit of algebraic manipulation, we can find two linearly independent, real-value solutions to [link] and express our general solution in those terms.

We encountered exponential functions with complex exponents earlier. One of the key tools we used to express these exponential functions in terms of sines and cosines was Euler’s formula , which tells us that

${e}^{i\theta }=\text{cos}\phantom{\rule{0.1em}{0ex}}\theta +i\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta$

for all real numbers $\theta \text{.}$

Going back to the general solution, we have

$\begin{array}{cc}\hfill y\left(x\right)& ={c}_{1}{e}^{\left(\alpha +\beta i\right)x}+{c}_{2}{e}^{\left(\alpha -\beta i\right)x}\hfill \\ & ={c}_{1}{e}^{\alpha x}{e}^{\beta ix}+{c}_{2}{e}^{\alpha x}{e}^{\text{−}\beta ix}\hfill \\ & ={e}^{\alpha x}\left({c}_{1}{e}^{\beta ix}+{c}_{2}{e}^{\text{−}\beta ix}\right).\hfill \end{array}$

Applying Euler’s formula together with the identities $\text{cos}\left(\text{−}x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x$ and $\text{sin}\left(\text{−}x\right)=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}x,$ we get

$\begin{array}{cc}\hfill y\left(x\right)& ={e}^{\alpha x}\left[{c}_{1}\left(\text{cos}\phantom{\rule{0.1em}{0ex}}\beta x+i\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\beta x\right)+{c}_{2}\left(\text{cos}\left(\text{−}\beta x\right)+i\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\text{−}\beta x\right)\right)\right]\hfill \\ & ={e}^{\alpha x}\left[\left({c}_{1}+{c}_{2}\right)\text{cos}\phantom{\rule{0.1em}{0ex}}\beta x+\left({c}_{1}-{c}_{2}\right)i\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\beta x\right].\hfill \end{array}$

Now, if we choose ${c}_{1}={c}_{2}=\frac{1}{2},$ the second term is zero and we get

$y\left(x\right)={e}^{\alpha x}\text{cos}\phantom{\rule{0.1em}{0ex}}\beta x$

as a real-value solution to [link] . Similarly, if we choose ${c}_{1}=\text{−}\frac{i}{2}$ and ${c}_{2}=\frac{i}{2},$ the first term is zero and we get

$y\left(x\right)={e}^{\alpha x}\text{sin}\phantom{\rule{0.1em}{0ex}}\beta x$

as a second, linearly independent, real-value solution to [link] .

Based on this, we see that if the characteristic equation has complex conjugate roots $\alpha ±\beta i,$ then the general solution to [link] is given by

$\begin{array}{cc}\hfill y\left(x\right)& ={c}_{1}{e}^{\alpha x}\text{cos}\phantom{\rule{0.1em}{0ex}}\beta x+{c}_{2}{e}^{\alpha x}\text{sin}\phantom{\rule{0.1em}{0ex}}\beta x\hfill \\ & ={e}^{\alpha x}\left({c}_{1}\text{cos}\phantom{\rule{0.1em}{0ex}}\beta x+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}\beta x\right),\hfill \end{array}$

where ${c}_{1}$ and ${c}_{2}$ are constants.

For example, the differential equation $y\text{″}-2{y}^{\prime }+5y=0$ has the associated characteristic equation ${\lambda }^{2}-2\lambda +5=0.$ By the quadratic formula, the roots of the characteristic equation are $1±2i\text{.}$ Therefore, the general solution to this differential equation is

$y\left(x\right)={e}^{x}\left({c}_{1}\text{cos}\phantom{\rule{0.1em}{0ex}}2x+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}2x\right).$

## Summary of results

We can solve second-order, linear, homogeneous differential equations with constant coefficients by finding the roots of the associated characteristic equation. The form of the general solution varies, depending on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots. The three cases are summarized in [link] .

Summary of characteristic equation cases
Characteristic Equation Roots General Solution to the Differential Equation
Distinct real roots, ${\lambda }_{1}$ and ${\lambda }_{2}$ $y\left(x\right)={c}_{1}{e}^{{\lambda }_{1}x}+{c}_{2}{e}^{{\lambda }_{2}x}$
A repeated real root, $\lambda$ $y\left(x\right)={c}_{1}{e}^{\lambda x}+{c}_{2}x{e}^{\lambda x}$
Complex conjugate roots $\alpha ±\beta i$ $y\left(x\right)={e}^{\alpha x}\left({c}_{1}\text{cos}\phantom{\rule{0.1em}{0ex}}\beta x+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}\beta x\right)$

## Problem-solving strategy: using the characteristic equation to solve second-order differential equations with constant coefficients

1. Write the differential equation in the form $ay\text{″}+b{y}^{\prime }+cy=0.$
2. Find the corresponding characteristic equation $a{\lambda }^{2}+b\lambda +c=0.$
3. Either factor the characteristic equation or use the quadratic formula to find the roots.
4. Determine the form of the general solution based on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots.

## Solving second-order equations with constant coefficients

Find the general solution to the following differential equations. Give your answers as functions of x .

1. $y\text{″}+3{y}^{\prime }-4y=0$
2. $y\text{″}+6{y}^{\prime }+13y=0$
3. $y\text{″}+2{y}^{\prime }+y=0$
4. $y\text{″}-5{y}^{\prime }=0$
5. $y\text{″}-16y=0$
6. $y\text{″}+16y=0$

Note that all these equations are already given in standard form (step 1).

1. The characteristic equation is ${\lambda }^{2}+3\lambda -4=0$ (step 2). This factors into $\left(\lambda +4\right)\left(\lambda -1\right)=0,$ so the roots of the characteristic equation are ${\lambda }_{1}=-4$ and ${\lambda }_{2}=1$ (step 3). Then the general solution to the differential equation is
$y\left(x\right)={c}_{1}{e}^{-4x}+{c}_{2}{e}^{x}\phantom{\rule{0.2em}{0ex}}\text{(step 4)}\text{.}$
2. The characteristic equation is ${\lambda }^{2}+6\lambda +13=0$ (step 2). Applying the quadratic formula, we see this equation has complex conjugate roots $-3±2i$ (step 3). Then the general solution to the differential equation is
$y\left(t\right)={e}^{-3t}\left({c}_{1}\text{cos}\phantom{\rule{0.1em}{0ex}}2t+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}2t\right)\phantom{\rule{0.2em}{0ex}}\text{(step 4)}\text{.}$
3. The characteristic equation is ${\lambda }^{2}+2\lambda +1=0$ (step 2). This factors into ${\left(\lambda +1\right)}^{2}=0,$ so the characteristic equation has a repeated real root $\lambda =-1$ (step 3). Then the general solution to the differential equation is
$y\left(t\right)={c}_{1}{e}^{\text{−}t}+{c}_{2}t{e}^{\text{−}t}\phantom{\rule{0.2em}{0ex}}\text{(step 4).}$
4. The characteristic equation is ${\lambda }^{2}-5\lambda$ (step 2). This factors into $\lambda \left(\lambda -5\right)=0,$ so the roots of the characteristic equation are ${\lambda }_{1}=0$ and ${\lambda }_{2}=5$ (step 3). Note that ${e}^{0x}={e}^{0}=1,$ so our first solution is just a constant. Then the general solution to the differential equation is
$y\left(x\right)={c}_{1}+{c}_{2}{e}^{5x}\phantom{\rule{0.2em}{0ex}}\text{(step 4)}\text{.}$
5. The characteristic equation is ${\lambda }^{2}-16=0$ (step 2). This factors into $\left(\lambda +4\right)\left(\lambda -4\right)=0,$ so the roots of the characteristic equation are ${\lambda }_{1}=4$ and ${\lambda }_{2}=-4$ (step 3). Then the general solution to the differential equation is
$y\left(x\right)={c}_{1}{e}^{4x}+{c}_{2}{e}^{-4x}\phantom{\rule{0.2em}{0ex}}\text{(step 4)}\text{.}$
6. The characteristic equation is ${\lambda }^{2}+16=0$ (step 2). This has complex conjugate roots $±4i$ (step 3). Note that ${e}^{0x}={e}^{0}=1,$ so the exponential term in our solution is just a constant. Then the general solution to the differential equation is
$y\left(t\right)={c}_{1}\text{cos}\phantom{\rule{0.1em}{0ex}}4t+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}4t\phantom{\rule{0.2em}{0ex}}\text{(step 4)}\text{.}$

Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!