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Visit this website that discusses second-order differential equations.

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

  1. ( y ) 2 y + 8 x 3 y = 0
  2. ( sin t ) y + cos t 3 t y = 0
  1. Nonlinear
  2. Linear, nonhomogeneous
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Later in this section, we will see some techniques for solving specific types of differential equations. Before we get to that, however, let’s get a feel for how solutions to linear differential equations behave. In many cases, solving differential equations depends on making educated guesses about what the solution might look like. Knowing how various types of solutions behave will be helpful.

Verifying a solution

Consider the linear, homogeneous differential equation

x 2 y x y 3 y = 0 .

Looking at this equation, notice that the coefficient functions are polynomials, with higher powers of x associated with higher-order derivatives of y . Show that y = x 3 is a solution to this differential equation.

Let y = x 3 . Then y = 3 x 2 and y = 6 x . Substituting into the differential equation, we see that

x 2 y x y 3 y = x 2 ( 6 x ) x ( 3 x 2 ) 3 ( x 3 ) = 6 x 3 3 x 3 3 x 3 = 0 .
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Show that y = 2 x 2 is a solution to the differential equation

1 2 x 2 y x y + y = 0 .
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Although simply finding any solution to a differential equation is important, mathematicians and engineers often want to go beyond finding one solution to a differential equation to finding all solutions to a differential equation. In other words, we want to find a general solution . Just as with first-order differential equations, a general solution (or family of solutions) gives the entire set of solutions to a differential equation. An important difference between first-order and second-order equations is that, with second-order equations, we typically need to find two different solutions to the equation to find the general solution. If we find two solutions, then any linear combination of these solutions is also a solution. We state this fact as the following theorem.

Superposition principle

If y 1 ( x ) and y 2 ( x ) are solutions to a linear homogeneous differential equation, then the function

y ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) ,

where c 1 and c 2 are constants, is also a solution.

The proof of this superposition principle theorem is left as an exercise.

Verifying the superposition principle

Consider the differential equation

y 4 y 5 y = 0 .

Given that e x and e 5 x are solutions to this differential equation, show that 4 e x + e 5 x is a solution.

We have

y ( x ) = 4 e x + e 5 x , so y ( x ) = −4 e x + 5 e 5 x and y ( x ) = 4 e x + 25 e 5 x .

Then

y 4 y 5 y = ( 4 e x + 25 e 5 x ) 4 ( −4 e x + 5 e 5 x ) 5 ( 4 e x + e 5 x ) = 4 e x + 25 e 5 x + 16 e x 20 e 5 x 20 e x 5 e 5 x = 0 .

Thus, y ( x ) = 4 e x + e 5 x is a solution.

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Consider the differential equation

y + 5 y + 6 y = 0 .

Given that e −2 x and e −3 x are solutions to this differential equation, show that 3 e −2 x + 6 e −3 x is a solution.

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Unfortunately, to find the general solution to a second-order differential equation, it is not enough to find any two solutions and then combine them. Consider the differential equation

x + 7 x + 12 x = 0 .

Both e −3 t and 2 e −3 t are solutions (check this). However, x ( t ) = c 1 e −3 t + c 2 ( 2 e −3 t ) is not the general solution. This expression does not account for all solutions to the differential equation. In particular, it fails to account for the function e −4 t , which is also a solution to the differential equation.

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
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Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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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.
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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