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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.

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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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