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  • Use separation of variables to solve a differential equation.
  • Solve applications using separation of variables.

We now examine a solution technique for finding exact solutions to a class of differential equations known as separable differential equations. These equations are common in a wide variety of disciplines, including physics, chemistry, and engineering. We illustrate a few applications at the end of the section.

Separation of variables

We start with a definition and some examples.


A separable differential equation    is any equation that can be written in the form

y = f ( x ) g ( y ) .

The term ‘separable’ refers to the fact that the right-hand side of the equation can be separated into a function of x times a function of y . Examples of separable differential equations include

y = ( x 2 4 ) ( 3 y + 2 ) y = 6 x 2 + 4 x y = sec y + tan y y = x y + 3 x 2 y 6.

The second equation is separable with f ( x ) = 6 x 2 + 4 x and g ( y ) = 1 , the third equation is separable with f ( x ) = 1 and g ( y ) = sec y + tan y , and the right-hand side of the fourth equation can be factored as ( x + 3 ) ( y 2 ) , so it is separable as well. The third equation is also called an autonomous differential equation    because the right-hand side of the equation is a function of y alone. If a differential equation is separable, then it is possible to solve the equation using the method of separation of variables    .

Problem-solving strategy: separation of variables

  1. Check for any values of y that make g ( y ) = 0 . These correspond to constant solutions.
  2. Rewrite the differential equation in the form d y g ( y ) = f ( x ) d x .
  3. Integrate both sides of the equation.
  4. Solve the resulting equation for y if possible.
  5. If an initial condition exists, substitute the appropriate values for x and y into the equation and solve for the constant.

Note that Step 4. states “Solve the resulting equation for y if possible.” It is not always possible to obtain y as an explicit function of x . Quite often we have to be satisfied with finding y as an implicit function of x .

Using separation of variables

Find a general solution to the differential equation y = ( x 2 4 ) ( 3 y + 2 ) using the method of separation of variables.

Follow the five-step method of separation of variables.

  1. In this example, f ( x ) = x 2 4 and g ( y ) = 3 y + 2 . Setting g ( y ) = 0 gives y = 2 3 as a constant solution.
  2. Rewrite the differential equation in the form
    d y 3 y + 2 = ( x 2 4 ) d x .
  3. Integrate both sides of the equation:
    d y 3 y + 2 = ( x 2 4 ) d x .

    Let u = 3 y + 2 . Then d u = 3 d y d x d x , so the equation becomes
    1 3 1 u d u = 1 3 x 3 4 x + C 1 3 ln | u | = 1 3 x 3 4 x + C 1 3 ln | 3 y + 2 | = 1 3 x 3 4 x + C .
  4. To solve this equation for y , first multiply both sides of the equation by 3 .
    ln | 3 y + 2 | = x 3 12 x + 3 C

    Now we use some logic in dealing with the constant C . Since C represents an arbitrary constant, 3 C also represents an arbitrary constant. If we call the second arbitrary constant C 1 , the equation becomes
    ln | 3 y + 2 | = x 3 12 x + C 1 .

    Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base e ) .
    e ln | 3 y + 2 | = e x 3 12 x + C 1 | 3 y + 2 | = e C 1 e x 3 12 x

    Again define a new constant C 2 = e c 1 (note that C 2 > 0 ) :
    | 3 y + 2 | = C 2 e x 3 12 x .

    This corresponds to two separate equations: 3 y + 2 = C 2 e x 3 12 x and 3 y + 2 = C 2 e x 3 12 x .
    The solution to either equation can be written in the form y = −2 ± C 2 e x 3 12 x 3 .
    Since C 2 > 0 , it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant C is entirely arbitrary, and can be dropped. Therefore the solution can be written as
    y = −2 + C e x 3 12 x 3 .
  5. No initial condition is imposed, so we are finished.
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Questions & Answers

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Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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