# 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
$y\left(t\right)={c}_{1}\text{cos}\phantom{\rule{0.1em}{0ex}}4t+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}4t\phantom{\rule{0.2em}{0ex}}\text{(step 4)}\text{.}$

#### Questions & Answers

where we get a research paper on Nano chemistry....?
Maira Reply
what are the products of Nano chemistry?
Maira Reply
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
Google
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
Jyoti Reply
I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
what is a peer
LITNING Reply
What is meant by 'nano scale'?
LITNING Reply
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Good
can you provide the details of the parametric equations for the lines that defince doubly-ruled surfeces (huperbolids of one sheet and hyperbolic paraboloid). Can you explain each of the variables in the equations?
Radek Reply

### Read also:

#### Get the best Algebra and trigonometry course in your pocket!

Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 3' conversation and receive update notifications?

 By Brooke Delaney By Richley Crapo By Rohini Ajay By By Steve Gibbs By Richley Crapo By OpenStax By Stephen Voron By Saylor Foundation By Rylee Minllic