# 7.1 Second-order linear equations  (Page 8/15)

 Page 8 / 15

Solve the following initial-value problem and graph the solution: $y\text{″}-2{y}^{\prime }+10y=0,y\left(0\right)=2,{y}^{\prime }\left(0\right)=-1$

$y\left(x\right)={e}^{x}\left(2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}3x-\text{sin}\phantom{\rule{0.1em}{0ex}}3x\right)$

## Initial-value problem representing a spring-mass system

The following initial-value problem models the position of an object with mass attached to a spring. Spring-mass systems are examined in detail in Applications . The solution to the differential equation gives the position of the mass with respect to a neutral (equilibrium) position (in meters) at any given time. (Note that for spring-mass systems of this type, it is customary to define the downward direction as positive.)

$y\text{″}+2{y}^{\prime }+y=0,y\left(0\right)=1,{y}^{\prime }\left(0\right)=0$

Solve the initial-value problem and graph the solution. What is the position of the mass at time $t=2$ sec? How fast is the mass moving at time $t=1$ sec? In what direction?

In [link] c. we found the general solution to this differential equation to be

$y\left(t\right)={c}_{1}{e}^{\text{−}t}+{c}_{2}t{e}^{\text{−}t}.$

Then

${y}^{\prime }\left(t\right)=\text{−}{c}_{1}{e}^{\text{−}t}+{c}_{2}\left(\text{−}t{e}^{\text{−}t}+{e}^{\text{−}t}\right).$

When $t=0,$ we have $y\left(0\right)={c}_{1}$ and ${y}^{\prime }\left(0\right)=\text{−}{c}_{1}+{c}_{2}.$ Applying the initial conditions, we obtain

$\begin{array}{ccc}\hfill {c}_{1}& =\hfill & 1\hfill \\ \hfill {\text{−}c}_{1}+{c}_{2}& =\hfill & 0.\hfill \end{array}$

Thus, ${c}_{1}=1,$ ${c}_{2}=1,$ and the solution to the initial value problem is

$y\left(t\right)={e}^{\text{−}t}+t{e}^{\text{−}t}.$

This solution is represented in the following graph. At time $t=2,$ the mass is at position $y\left(2\right)={e}^{-2}+2{e}^{-2}=3{e}^{-2}\approx 0.406$ m below equilibrium.

To calculate the velocity at time $t=1,$ we need to find the derivative. We have $y\left(t\right)={e}^{\text{−}t}+t{e}^{\text{−}t},$ so

${y}^{\prime }\left(t\right)=\text{−}{e}^{\text{−}t}+{e}^{\text{−}t}-t{e}^{\text{−}t}=\text{−}t{e}^{\text{−}t}.$

Then ${y}^{\prime }\left(1\right)=\text{−}{e}^{-1}\approx -0.3679.$ At time $t=1,$ the mass is moving upward at 0.3679 m/sec.

Suppose the following initial-value problem models the position (in feet) of a mass in a spring-mass system at any given time. Solve the initial-value problem and graph the solution. What is the position of the mass at time $t=0.3$ sec? How fast is it moving at time $t=0.1$ sec? In what direction?

$y\text{″}+14{y}^{\prime }+49y=0,y\left(0\right)=0,{y}^{\prime }\left(0\right)=1$

$y\left(t\right)=t{e}^{-7t}$

At time $t=0.3,$ $y\left(0.3\right)=0.3{e}^{\left(-7*0.3\right)}=0.3{e}^{-2.1}\approx 0.0367.$ The mass is 0.0367 ft below equilibrium. At time $t=0.1,$ ${y}^{\prime }\left(0.1\right)=0.3{e}^{-0.7}\approx 0.1490.$ The mass is moving downward at a speed of 0.1490 ft/sec.

## Solving a boundary-value problem

In [link] f. we solved the differential equation $y\text{″}+16y=0$ and found the general solution to be $y\left(t\right)={c}_{1}\text{cos}\phantom{\rule{0.1em}{0ex}}4t+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}4t\text{.}$ If possible, solve the boundary-value problem if the boundary conditions are the following:

1. $y\left(0\right)=0,$ $y\left(\frac{\pi }{4}\right)=0$
2. $y\left(0\right)=1,$ $y\left(\frac{\pi }{8}\right)=0$
3. $y\left(\frac{\pi }{8}\right)=0,$ $y\left(\frac{3\pi }{8}\right)=2$

We have

$y\left(x\right)={c}_{1}\text{cos}\phantom{\rule{0.1em}{0ex}}4t+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}4t\text{.}$
1. Applying the first boundary condition given here, we get $y\left(0\right)={c}_{1}=0.$ So the solution is of the form $y\left(t\right)={c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}4t\text{.}$ When we apply the second boundary condition, though, we get $y\left(\frac{\pi }{4}\right)={c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(4\left(\frac{\pi }{4}\right)\right)={c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}\pi =0$ for all values of ${c}_{2}.$ The boundary conditions are not sufficient to determine a value for ${c}_{2},$ so this boundary-value problem has infinitely many solutions. Thus, $y\left(t\right)={c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}4t$ is a solution for any value of ${c}_{2}.$
2. Applying the first boundary condition given here, we get $y\left(0\right)={c}_{1}=1.$ Applying the second boundary condition gives $y\left(\frac{\pi }{8}\right)={c}_{2}=0,$ so ${c}_{2}=0.$ In this case, we have a unique solution: $y\left(t\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}4t\text{.}$
3. Applying the first boundary condition given here, we get $y\left(\frac{\pi }{8}\right)={c}_{2}=0.$ However, applying the second boundary condition gives $y\left(\frac{3\pi }{8}\right)=\text{−}{c}_{2}=2,$ so ${c}_{2}=-2.$ We cannot have ${c}_{2}=0=-2,$ so this boundary value problem has no solution.

## Key concepts

• Second-order differential equations can be classified as linear or nonlinear, homogeneous or nonhomogeneous.
• To find a general solution for a homogeneous second-order differential equation, we must find two linearly independent solutions. If ${y}_{1}\left(x\right)$ and ${y}_{2}\left(x\right)$ are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by
$y\left(x\right)={c}_{1}{y}_{1}\left(x\right)+{c}_{2}{y}_{2}\left(x\right).$
• To solve homogeneous second-order differential equations with constant coefficients, find the roots of the 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.
• Initial conditions or boundary conditions can then be used to find the specific solution to a differential equation that satisfies those conditions, except when there is no solution or infinitely many solutions.

## Key equations

• Linear second-order differential equation
${a}_{2}\left(x\right)y\text{″}+{a}_{1}\left(x\right){y}^{\prime }+{a}_{0}\left(x\right)y=r\left(x\right)$
• Second-order equation with constant coefficients
$ay\text{″}+b{y}^{\prime }+cy=0$

Classify each of the following equations as linear or nonlinear. If the equation is linear, determine whether it is homogeneous or nonhomogeneous.

${x}^{3}y\text{″}+\left(x-1\right){y}^{\prime }-8y=0$

linear, homogenous

$\left(1+{y}^{2}\right)y\text{″}+x{y}^{\prime }-3y=\text{cos}\phantom{\rule{0.1em}{0ex}}x$

$xy\text{″}+{e}^{y}{y}^{\prime }=x$

nonlinear

$y\text{″}+\frac{4}{x}{y}^{\prime }-8xy=5{x}^{2}+1$

$y\text{″}+\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right){y}^{\prime }-xy=4y$

linear, homogeneous

$y\text{″}+\left(\frac{x+3}{y}\right){y}^{\prime }=0$

For each of the following problems, verify that the given function is a solution to the differential equation. Use a graphing utility to graph the particular solutions for several values of c 1 and c 2 . What do the solutions have in common?

[T] $y\text{″}+2{y}^{\prime }-3y=0;$ $y\left(x\right)={c}_{1}{e}^{x}+{c}_{2}{e}^{-3x}$

[T] ${x}^{2}y\text{″}-2y-3{x}^{2}+1=0;$ $y\left(x\right)={c}_{1}{x}^{2}+{c}_{2}{x}^{-1}+{x}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(x\right)+\frac{1}{2}$

[T] $y\text{″}+14{y}^{\prime }+49y=0;$ $y\left(x\right)={c}_{1}{e}^{-7x}+{c}_{2}x{e}^{-7x}$

[T] $6y\text{″}-49{y}^{\prime }+8y=0;$ $y\left(x\right)={c}_{1}{e}^{x\text{/}6}+{c}_{2}{e}^{8x}$

Find the general solution to the linear differential equation.

$y\text{″}-3{y}^{\prime }-10y=0$

$y={c}_{1}{e}^{5x}+{c}_{2}{e}^{-2x}$

$y\text{″}-7{y}^{\prime }+12y=0$

$y\text{″}+4{y}^{\prime }+4y=0$

$y={c}_{1}{e}^{-2x}+{c}_{2}x{e}^{-2x}$

$4y\text{″}-12{y}^{\prime }+9y=0$

$2y\text{″}-3{y}^{\prime }-5y=0$

$y={c}_{1}{e}^{5x\text{/}2}+{c}_{2}{e}^{\text{−}x}$

$3y\text{″}-14{y}^{\prime }+8y=0$

$y\text{″}+{y}^{\prime }+y=0$

$y={e}^{\text{−}x\text{/}2}\left({c}_{1}\text{cos}\phantom{\rule{0.1em}{0ex}}\frac{\sqrt{3}x}{2}+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}\frac{\sqrt{3}x}{2}\right)$

$5y\text{″}+2{y}^{\prime }+4y=0$

$y\text{″}-121y=0$

$y={c}_{1}{e}^{-11x}+{c}_{2}{e}^{11x}$

$8y\text{″}+14{y}^{\prime }-15y=0$

$y\text{″}+81y=0$

$y={c}_{1}\text{cos}\phantom{\rule{0.1em}{0ex}}9x+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}9x$

$y\text{″}-{y}^{\prime }+11y=0$

$2y\text{″}=0$

$y={c}_{1}+{c}_{2}x$

$y\text{″}-6{y}^{\prime }+9y=0$

$3y\text{″}-2{y}^{\prime }-7y=0$

$y={c}_{1}{e}^{\left(\left(1+\sqrt{22}\right)\text{/}3\right)x}+{c}_{2}{e}^{\left(\left(1-\sqrt{22}\right)\text{/}3\right)x}$

$4y\text{″}-10{y}^{\prime }=0$

$36\frac{{d}^{2}y}{d{x}^{2}}+12\frac{dy}{dx}+y=0$

$y={c}_{1}{e}^{\text{−}x\text{/}6}+{c}_{2}x{e}^{\text{−}x\text{/}6}$

$25\frac{{d}^{2}y}{d{x}^{2}}-80\frac{dy}{dx}+64y=0$

$\frac{{d}^{2}y}{d{x}^{2}}-9\frac{dy}{dx}=0$

$y={c}_{1}+{c}_{2}{e}^{9x}$

$4\frac{{d}^{2}y}{d{x}^{2}}+8y=0$

Solve the initial-value problem.

$y\text{″}+5{y}^{\prime }+6y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=0,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(0\right)=-2$

$y=-2{e}^{-2x}+2{e}^{-3x}$

$y\text{″}+2{y}^{\prime }-8y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=5,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(0\right)=4$

$y\text{″}+4y=0,\phantom{\rule{4em}{0ex}}y\left(0\right)=3,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(0\right)=10$

$y=3\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(2x\right)+5\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(2x\right)$

$y\text{″}-18{y}^{\prime }+81y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=1,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(0\right)=5$

$y\text{″}-{y}^{\prime }-30y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=1,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(0\right)=-16$

$y=\text{−}{e}^{6x}+2{e}^{-5x}$

$4y\text{″}+4{y}^{\prime }-8y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=2,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(0\right)=1$

$25y\text{″}+10{y}^{\prime }+y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=2,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(0\right)=1$

$y=2{e}^{\text{−}x\text{/}5}+\frac{7}{5}x{e}^{\text{−}x\text{/}5}$

$y\text{″}+y=0,\phantom{\rule{4em}{0ex}}y\left(\pi \right)=1,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(\pi \right)=-5$

Solve the boundary-value problem, if possible.

$y\text{″}+{y}^{\prime }-42y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=0,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(1\right)=2$

$y=\left(\frac{2}{{e}^{6}-{e}^{-7}}\right){e}^{6x}-\left(\frac{2}{{e}^{6}-{e}^{-7}}\right){e}^{-7x}$

$9y\text{″}+y=0,\phantom{\rule{4em}{0ex}}y\left(\frac{3\pi }{2}\right)=6,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(0\right)=-8$

$y\text{″}+10{y}^{\prime }+34y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=6,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(\pi \right)=2$

No solutions exist.

$y\text{″}+7{y}^{\prime }-60y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=4,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(2\right)=0$

$y\text{″}-4{y}^{\prime }+4y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=2,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(1\right)=-1$

$y=2{e}^{2x}-\frac{2{e}^{2}+1}{{e}^{2}}x{e}^{2x}$

$y\text{″}-5{y}^{\prime }=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=3,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(-1\right)=2$

$y\text{″}+9y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=4,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(\frac{\pi }{3}\right)=-4$

$y=4\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}3x+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}3x,\phantom{\rule{0.2em}{0ex}}\text{infinitely many solutions}$

$4y\text{″}+25y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=2,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(2\pi \right)=-2$

Find a differential equation with a general solution that is $y={c}_{1}{e}^{x\text{/}5}+{c}_{2}{e}^{-4x}.$

$5y\text{″}+19{y}^{\prime }-4y=0$

Find a differential equation with a general solution that is $y={c}_{1}{e}^{x}+{c}_{2}{e}^{-4x\text{/}3}.$

For each of the following differential equations:

1. Solve the initial value problem.
2. [T] Use a graphing utility to graph the particular solution.

$y\text{″}+64y=0;\phantom{\rule{2em}{0ex}}y\left(0\right)=3,\phantom{\rule{1em}{0ex}}{y}^{\prime }\left(0\right)=16$

a. $y=3\phantom{\rule{0.1em}{0ex}}\text{cos}\left(8x\right)+2\phantom{\rule{0.1em}{0ex}}\text{sin}\left(8x\right)$
b.

$y\text{″}-2{y}^{\prime }+10y=0\phantom{\rule{2em}{0ex}}y\left(0\right)=1,\phantom{\rule{1em}{0ex}}{y}^{\prime }\left(0\right)=13$

$y\text{″}+5{y}^{\prime }+15y=0\phantom{\rule{2em}{0ex}}y\left(0\right)=-2,\phantom{\rule{1em}{0ex}}{y}^{\prime }\left(0\right)=7$

a. $y={e}^{\left(-5\text{/}2\right)x}\left[-2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(\frac{\sqrt{35}}{2}x\right)+\frac{4\sqrt{35}}{35}\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(\frac{\sqrt{35}}{2}x\right)\right]$
b.

(Principle of superposition) Prove that if ${y}_{1}\left(x\right)$ and ${y}_{2}\left(x\right)$ are solutions to a linear homogeneous differential equation, $y\text{″}+p\left(x\right){y}^{\prime }+q\left(x\right)y=0,$ then the function $y\left(x\right)={c}_{1}{y}_{1}\left(x\right)+{c}_{2}{y}_{2}\left(x\right),$ where ${c}_{1}$ and ${c}_{2}$ are constants, is also a solution.

Prove that if a, b, and c are positive constants, then all solutions to the second-order linear differential equation $ay\text{″}+b{y}^{\prime }+cy=0$ approach zero as $x\to \infty \text{.}$ ( Hint: Consider three cases: two distinct roots, repeated real roots, and complex conjugate roots.)

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
Berger describes sociologists as concerned with
Got questions? Join the online conversation and get instant answers!