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Examples of first-order nonlinear differential equations include

( y ) 4 ( y ) 3 = ( 3 x 2 ) ( y + 4 ) 4 y + 3 y 3 = 4 x 5 ( y ) 2 = sin y + cos x .

These equations are nonlinear because of terms like ( y ) 4 , y 3 , etc. Due to these terms, it is impossible to put these equations into the same form as [link] .

Standard form

Consider the differential equation

( 3 x 2 4 ) y + ( x 3 ) y = sin x .

Our main goal in this section is to derive a solution method for equations of this form. It is useful to have the coefficient of y be equal to 1 . To make this happen, we divide both sides by 3 x 2 4 .

y + ( x 3 3 x 2 4 ) y = sin x 3 x 2 4

This is called the standard form    of the differential equation. We will use it later when finding the solution to a general first-order linear differential equation. Returning to [link] , we can divide both sides of the equation by a ( x ) . This leads to the equation

y + b ( x ) a ( x ) y = c ( x ) a ( x ) .

Now define p ( x ) = b ( x ) a ( x ) and q ( x ) = c ( x ) a ( x ) . Then [link] becomes

y + p ( x ) y = q ( x ) .

We can write any first-order linear differential equation in this form, and this is referred to as the standard form for a first-order linear differential equation.

Writing first-order linear equations in standard form

Put each of the following first-order linear differential equations into standard form. Identify p ( x ) and q ( x ) for each equation.

  1. y = 3 x 4 y
  2. 3 x y 4 y 3 = 2 (here x > 0 )
  3. y = 3 y 4 x 2 + 5
  1. Add 4 y to both sides:
    y + 4 y = 3 x .

    In this equation, p ( x ) = 4 and q ( x ) = 3 x .
  2. Multiply both sides by 4 y 3 , then subtract 8 y from each side:
    3 x y 4 y 3 = 2 3 x y = 2 ( 4 y 3 ) 3 x y = 8 y 6 3 x y 8 y = −6 .

    Finally, divide both sides by 3 x to make the coefficient of y equal to 1 :
    y 8 3 x y = 2 3 x .
    This is allowable because in the original statement of this problem we assumed that x > 0 . (If x = 0 then the original equation becomes 0 = 2 , which is clearly a false statement.)
    In this equation, p ( x ) = 8 3 x and q ( x ) = 2 3 x .
  3. Subtract y from each side and add 4 x 2 5 :
    3 y y = 4 x 2 5 .

    Next divide both sides by 3 :
    y 1 3 y = 4 3 x 2 5 3 .

    In this equation, p ( x ) = 1 3 and q ( x ) = 4 3 x 2 5 3 .
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Put the equation ( x + 3 ) y 2 x 3 y 4 = 5 into standard form and identify p ( x ) and q ( x ) .

y + 15 x + 3 y = 10 x 20 x + 3 ; p ( x ) = 15 x + 3 and q ( x ) = 10 x 20 x + 3

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Integrating factors

We now develop a solution technique for any first-order linear differential equation. We start with the standard form of a first-order linear differential equation:

y + p ( x ) y = q ( x ) .

The first term on the left-hand side of [link] is the derivative of the unknown function, and the second term is the product of a known function with the unknown function. This is somewhat reminiscent of the power rule from the Differentiation Rules section. If we multiply [link] by a yet-to-be-determined function μ ( x ) , then the equation becomes

μ ( x ) y + μ ( x ) p ( x ) y = μ ( x ) q ( x ) .

The left-hand side [link] can be matched perfectly to the product rule:

d d x [ f ( x ) g ( x ) ] = f ( x ) g ( x ) + f ( x ) g ( x ) .

Matching term by term gives y = f ( x ) , g ( x ) = μ ( x ) , and g ( x ) = μ ( x ) p ( x ) . Taking the derivative of g ( x ) = μ ( x ) and setting it equal to the right-hand side of g ( x ) = μ ( x ) p ( x ) leads to

μ ( x ) = μ ( x ) p ( x ) .

This is a first-order, separable differential equation for μ ( x ) . We know p ( x ) because it appears in the differential equation we are solving. Separating variables and integrating yields

μ ( x ) μ ( x ) = p ( x ) μ ( x ) μ ( x ) d x = p ( x ) d x ln | μ ( x ) | = p ( x ) d x + C e ln | μ ( x ) | = e p ( x ) d x + C | μ ( x ) | = C 1 e p ( x ) d x μ ( x ) = C 2 e p ( x ) d x .

Questions & Answers

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Ali
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The nanotechnology is as new science, to scale nanometric
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nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
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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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You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?
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Practice Key Terms 3

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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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