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
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?
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what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.?
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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
The nanotechnology is as new science, to scale nanometric
brayan
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Damian
Is there any normative that regulates the use of silver nanoparticles?
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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?