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
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.
In this morden time nanotechnology used in many field .
1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc
2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc
3- Atomobile -MEMS, Coating on car etc.
and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change .
maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.