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
Applying Euler’s formula together with the identities
$\text{cos}(\text{\u2212}x)=\text{cos}\phantom{\rule{0.1em}{0ex}}x$ and
$\text{sin}(\text{\u2212}x)=\text{\u2212}\text{sin}\phantom{\rule{0.1em}{0ex}}x,$ we get
as a real-value solution to
[link] . Similarly, if we choose
${c}_{1}=\text{\u2212}\frac{i}{2}$ and
${c}_{2}=\frac{i}{2},$ the first term is zero and we get
as a second, linearly independent, real-value solution to
[link] .
Based on this, we see that if the characteristic equation has complex conjugate roots
$\alpha \pm \beta i,$ then the general solution to
[link] is given by
For example, the differential equation
$y\text{\u2033}-2{y}^{\prime}+5y=0$ has the associated characteristic equation
${\lambda}^{2}-2\lambda +5=0.$ By the quadratic formula, the roots of the characteristic equation are
$1\pm 2i\text{.}$ Therefore, the general solution to this differential equation is
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
${\lambda}_{2}$
Problem-solving strategy: using the characteristic equation to solve second-order differential equations with constant coefficients
Write the differential equation in the form
$ay\text{\u2033}+b{y}^{\prime}+cy=0.$
Find the corresponding characteristic equation
$a{\lambda}^{2}+b\lambda +c=0.$
Either factor the characteristic equation or use the quadratic formula to find the roots.
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 .
$y\text{\u2033}+3{y}^{\prime}-4y=0$
$y\text{\u2033}+6{y}^{\prime}+13y=0$
$y\text{\u2033}+2{y}^{\prime}+y=0$
$y\text{\u2033}-5{y}^{\prime}=0$
$y\text{\u2033}-16y=0$
$y\text{\u2033}+16y=0$
Note that all these equations are already given in standard form (step 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
${\lambda}_{1}=\mathrm{-4}$ and
${\lambda}_{2}=1$ (step 3). Then the general solution to the differential equation is
The characteristic equation is
${\lambda}^{2}+6\lambda +13=0$ (step 2). Applying the quadratic formula, we see this equation has complex conjugate roots
$\mathrm{-3}\pm 2i$ (step 3). Then the general solution to the differential equation is
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
$\lambda =\mathrm{-1}$ (step 3). Then the general solution to the differential equation is
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
The characteristic equation is
${\lambda}^{2}-16=0$ (step 2). This factors into
$(\lambda +4)(\lambda -4)=0,$ so the roots of the characteristic equation are
${\lambda}_{1}=4$ and
${\lambda}_{2}=\mathrm{-4}$ (step 3). Then the general solution to the differential equation is
The characteristic equation is
${\lambda}^{2}+16=0$ (step 2). This has complex conjugate roots
$\pm 4i$ (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
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
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?