2.5 Solutions by Substitutions

INTRODUCTION

We usually solve a differential equation by recognizing it as a certain kind of equation (say, separable) and then carrying out a procedure, consisting of equation-specific mathematical steps, that yields a function that satisfies the equation. Often the first step in solving a given differential equation consists of transforming it into another differential equation by means of a substitution. For example, suppose we wish to transform the first-order equation dy/dx = f(x, y) by the substitution y = g(x, u), where u is regarded as a function of the variable x.

If g possesses first-partial derivatives, then the Chain Rule gives

By replacing dy/dx by f(x, y) and y by g(x, u) in the foregoing derivative, we get the new first-order differential equation

which, after solving for du/dx, has the form du/dx = F(x, u). If we can determine a solution u = ϕ(x) of this second equation, then a solution of the original differential equation is y = g(x, ϕ(x)).

Homogeneous Equations

If a function f possesses the property for some real number α, then f is said to be a homogeneous function of degree α. For example, f(x, y) = x3 + y3 is a homogeneous function of degree 3 since

f(tx, ty) = (tx)3 + (ty)3 = t3(x3 + y3) = t3f(x, y),

whereas f(x, y) = x3 + y3 + 1 is seen not to be homogeneous. The functions

are homogeneous of degrees 0 and −1, respectively, because

A first-order DE written in differential form

(1)

is said to be a homogeneous equation if the coefficient functions M(x, y) and N(x, y) are homogeneous functions of the same degree. In other words, (1) is a homogeneous equation if

M(tx, ty) = tαM(x, y)     and     N(tx, ty) = tαN(x, y).

A linear first-order DE a1y′ + a0y = g(x) is homogeneous when g(x) = 0.

The word homogeneous as used here does not mean the same as it does when applied to linear differential equations. See Sections 2.3 and 3.1.

If M and N are homogeneous functions of degree α, we can also write

M(x, y) = xαM(1, u)     and     N(x, y) = xαN(1, u)     where u = y/x,(2)

and M(x, y) = yαM(v, 1)     and     N(x, y) = yαN(v, 1)     where v = x/y.(3)

See Problem 33 in Exercises 2.5. Properties (2) and (3) suggest the substitutions that can be used to solve a homogeneous differential equation. Specifically, either of the substitutions , where u and v are new dependent variables, will reduce a homogeneous equation to a separable first-order differential equation. To show this, observe that as a consequence of (2) a homogeneous equation M(x, y) dx + N(x, y) dy = 0 can be rewritten as

xα M(1, u) dx + xα N(1, u) dy = 0     or     M(1, u) dx + N(1, u) dy = 0,

where u = y/x or y = ux. By substituting the differential dy = u dx + x du into the last equation and gathering terms, we obtain a separable DE in the variables u and x:

or

We hasten to point out that the preceding formula should not be memorized; rather, the procedure should be worked through each time. The proof that the substitutions x = vy and dx = v dy + y dv also lead to a separable equation follows in an analogous manner from (3).

EXAMPLE 1 Solving a Homogeneous DE

Solve (x2 + y2) dx + (x2 − xy) dy = 0.

SOLUTION

Inspection of M(x, y) = x2 + y2 and N(x, y) = x2 − xy shows that these coefficients are homogeneous functions of degree 2. If we let y = ux, then dy = u dx + x du so that, after substituting, the given equation becomes

After integration the last line gives

Using the properties of logarithms, we can write the preceding solution as

    or     ≡

Although either of the indicated substitutions can be used for every homogeneous differential equation, in practice we try x = vy whenever the function M(x, y) is simpler than N(x, y). Also it could happen that after using one substitution, we may encounter integrals that are difficult or impossible to evaluate in closed form; switching substitutions may result in an easier problem.

Bernoulli’s Equation

The differential equation

,(4)

where n is any real number, is called Bernoulli’s equation after the Swiss mathematician Jacob Bernoulli (1654–1705), who studied the equation around 1695. Note that for n = 0 equation (4) is linear, and for n = 1 equation (4) is separable and linear. In 1696 the German polymath Gottfried Wilhelm Leibniz (1646–1716) showed that for n ≠ 0 and n ≠ 1, any differential equation of the form given in (4) can be reduced to a linear equation by means of the substitution . The next example illustrates this substitution method.

EXAMPLE 2 Solving a Bernoulli DE

Solve

SOLUTION

We begin by rewriting the differential equation in the form given in (4) by dividing by x:

.

With n = 2 and u = y−1, we next substitute y = u–1 and

into the given equation and simplify. The result is

The integrating factor for this linear equation on, say, (0, ) is

gives x–1u = −x + c or u = −x2 + cx. Since u = y–1 we have y = 1/u, and so a solution of the given equation is . ≡

Note that we have not obtained the general solution of the original nonlinear differential equation in Example 2, since y = 0 is a singular solution of the equation.

Reduction to Separation of Variables

A differential equation of the form

(5)

can always be reduced to an equation with separable variables by means of the substitution u = Ax + By + C, B ≠ 0. Example 3 illustrates the technique.

EXAMPLE 3 An Initial-Value Problem

Solve the initial-value problem      = (−2x − y)2 + 7, y(0) = 0.

SOLUTION

If we let u = −2x + y, then du/dx = −2 + dy/dx, and so the differential equation is transformed into

    or    

The last equation is separable. Using partial fractions,

    or    

and integrating, then yields

    or    

Solving the last equation for u and then resubstituting gives the solution

    or     (6)

Finally, applying the initial condition y(0) = 0 to the last equation in (6) gives c = −1. With the aid of a graphing utility we have shown in FIGURE 2.5.1 the graph of the particular solution

in blue along with the graphs of some other members of the family solutions (6). ≡

The solution curve, marked in blue, begins at approximately (negative 1.3, 0.5), goes up and to the right to a high point at approximately (negative 0.5, 1.7), then goes down and to the right, passing through the origin. The curve continues into the fourth quadrant and reaches a low point at approximately (0.4, negative 1.7), then goes up and to the right to approximately (1.7, negative 0.5). 6 curves following the same trend are marked above and below the blue curve. The solution curves also have a branch that extends up and to the right with increasing steepness on the left, and a branch that goes down and to the right with decreasing steepness on the right.

FIGURE 2.5.1 Some solutions of the DE in Example 3

2.5 Exercises Answers to selected odd-numbered problems begin on page ANS-3.

Each DE in Problems 1–14 is homogeneous.

In Problems 1–10, solve the given differential equation by using an appropriate substitution.

  1. (x − y) dx + x dy = 0
  2. (x + y) dx + x dy = 0
  3. x dx + (y − 2x) dy = 0
  4. y dx = 2(x + y) dy
  5. (y2 + yx) dx − x2 dy = 0
  6. (y2 + yx) dx + x2 dy = 0

In Problems 11–14, solve the given initial-value problem.

  3. (x + yey/x) dx − xey/x dy = 0, y(1) = 0
  4. y dx + x(ln x − ln y − 1) dy = 0, y(1) = e

Each DE in Problems 15–22 is a Bernoulli equation.

In Problems 15–20, solve the given differential equation by using an appropriate substitution.

In Problems 21 and 22, solve the given initial-value problem.

  1. ,
  2. ,

Each DE in Problems 23–30 is of the form given in (5).

In Problems 23–28, solve the given differential equation by using an appropriate substitution.

In Problems 29 and 30, solve the given initial-value problem.

  1. ,     y(0) = π/4
  2. ,     

Discussion Problems

Some second-order differential equations can be reduced to a first-order equation by means of a substitution. In Problems 31 and 32, use the substitution u = y′ to reduce the given differential equation to a first-order equation, and then solve that equation by a method discussed in this section. Then use u = y′ again to find the solution y of the given equation.

  1. y″ + (y′)2 = 0
  2. y″ − 2y′ = 4x

  3. (a) Show that it is always possible to express any homogeneous differential equation M(x, y) dx + N(x, y) dy = 0 in the form

    (b) Show that the substitution u = y/x in the last differential equation in part (a) yields a separable equation.

  4. Put the homogeneous differential equation

    (5x2 + 2y2) dx − xy dy = 0

    in the form dy/dx = F(y/x) given in part (a) of Problem 33. Then solve the equation using the substitution u = y/x.

  5. (a) Determine two singular solutions of the DE in Problem 10.

    (b) If the initial condition y(5) = 0 is as prescribed in Problem 10, then what is the largest interval I over which the solution is defined? Use a graphing utility to plot the solution curve for the IVP.

  6. In Example 3, the solution y(x) becomes unbounded as x → ±. Nevertheless y(x) is asymptotic to a curve as x → − and to a different curve as x → . Find the equations of these curves.
  7. The differential equation

    is called a Riccati equation after the Venetian jurist and mathematician Jacopo Francesco Riccati (1676–1754) who was able to solve several DEs of this type.

    (a) A Riccati equation can be solved by a succession of two substitutions provided we know a particular solution y1 of the equation. Show that the substitution y = y1 + u reduces Riccati’s equation to a Bernoulli equation (4) with n = 2. The Bernoulli equation can then be reduced to a linear equation by the substitution w = u–1.

    (b) Find a one-parameter family of solutions for the differential equation

    where y1 = 2/x is a known solution of the equation.

  8. Devise an appropriate substitution to solve

    xy′ = y ln(xy).

Mathematical Models

  1. Population Growth     In the study of population dynamics one of the most famous models for a growing but bounded population is the logistic equation

    where a and b are positive constants. Although we will come back to this equation and solve it by an alternative method in Section 2.8, solve the DE this first time using the fact that it is a Bernoulli equation.

  2. Falling Chain     In Problem 47 in Exercises 2.4 we saw that a mathematical model for the velocity v of a chain slipping off the edge of a high horizontal platform is

    In that problem you were asked to solve the DE by converting it into an exact equation using an integrating factor. This time solve the DE using the fact that it is a Bernoulli equation.