Differential equations of the first order and first degree. Methods of solution. Separation of variables. Homogeneous, exact and linear equations. Integrating factors. Bernoulli�s equation.

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Differential equations of the first order and first degree. Methods of solution. Separation of variables. Homogeneous, exact and linear equations. Integrating factors. Bernoulli’s equation.

Differential equations of the first order and first degree. Any differential equation of the first order and first degree can be written in the form

Example. The differential equation

can also be written as

(x - 3y)dx + (x - 2y)dy = 0

Existence of a solution. The general solution of the equation dy/dx = g(x, y), if it exists, has the form f(x, y, C) = 0, where C is an arbitrary constant. Under what circumstances does a general solution exist? We have the following theorem.

Theorem 1. A general solution of dy/dx = g(x, y) exists over some specified region R of points (x, y) if the following conditions are met:

a) g(x, y) is continuous and single-valued over R

b) ∂g/∂y exists and is continuous at all points of R

The general solution f(x, y, C) = 0 of a differential equation dy/dx = g(x, y) over some region R consists of a family of curves, called the integral curves of the differential equation, (one curve for each possible value of C, each curve representing a particular solution), such that through each point in R there passes one and only one curve of the family f(x, y, C) = 0.

The differential equation

associates with each point (x₀, y₀) in the region R a direction

The direction at each point of R is that of the tangent to that curve of the family f(x, y, C) = 0 that passes through the point.

A region R in which a direction is associated with each point is called a direction field. In Fig. 1 is shown the direction field and integral curves for the differential equation dy/dx = 2x. The general solution of this equation is y = x² + C. The integral curves are parabolas.

Methods of solving differential equations of the first order and first degree

I Separation of variables. If an equation

M(x, y) dx + N(x, y) dy = 0

can be brought into the form

P(x) dx + Q(y) dy = 0 ,

the variables are referred to as having been separated. The general solution is then given by

Example. Solve

(1 + x²)dy - xy dx = 0

Solution. Dividing by y(1 + x²) and transposing we get

Integrating both sides, we get

ln y = ½ ln (1 + x²) + ln C

ln y = ln C(1 + x²)^½

Taking exponentials

y = C(1 + x²)^½

The arbitrary constant was added in the form “ln C” to facilitate the final representation.

Homogeneous polynomials, functions and equations

Def. Homogeneous polynomial. A polynomial whose terms are all of the same degree with respect to all the variables taken together. Thus

x² + 2xy - 2y² is homogeneous of degree 2

2x³y + 3 x²y² + 5y⁴ is homogeneous of degree 4

2x + 5y is homogeneous of degree 1

The concept of homogeneity is extended to general functions in the following way:

Def. Homogeneous function. A function such that if each of the variables is replaced by k times the variable, k can be completely factored out of the function whenever k ≠ 0. The power of k which can be factored out of the function is the degree of homogeneity of the function. Thus

2x²ln x/y + 4y² is homogeneous of degree 2

x²y + y³ sin y/x is homogeneous of degree 3

II Homogeneous differential equations. A differential equation of the form

M(x, y) dx + N(x, y) dy = 0

is said to be homogeneous if M(x, y) and N(x, y) are homogeneous functions of the same degree. Such an equation can be transformed into an equation in which the variables are separated by the substitution y = vx (or x = vy), where v is a new variable.

Note. Differentiating y = vx gives dy = v dx + x dv, a quantity that must be substituted for dy when vx is substituted for y.

Example. Solve

(x² - y²)dx + 2xy dy = 0

Solution. We cannot separate the variables, but M(x, y) and N(x, y) are homogeneous functions of degree 2. Substituting

y = vx and dy = v dx + x dv

we get

(1 - v²)dx + 2v(v dx + x dv) = 0

Separating the variables gives

Integrating we get

ln(v² + 1) = - ln x + ln C

Taking exponentials we obtain

x(v² + 1) = C

Finally, since v = y/x, this becomes

x² + y² = Cx

Reason why the substitution y = vx transforms the equation into one in which the variables are separable. The reason the substitution y = vx transforms the equation into one in which the variables are separable can be seen when the given equation is written in the form

If M(x, y) and N(x, y) are homogeneous functions of the same degree and one substitutes vx for y one finds that the x’s all cancel out on the right side of 2) and the right side becomes a function in v alone i.e. the equation takes the form

3) dy/dx = g(v)

Substituting dy = v dx + x dv then gives

4) v dx + x dv = g(v) dx

where the variables can be separated as

III Exact differential equations

Def. Exact differential equation. A differential equation which is obtained by setting the total differential of some function equal to zero.

The total differential of a function u(x, y) is, by definition,

and the exact differential equation associated with the function u(x, y) is

The primitive of 5) is u(x, y) = C.

If in a given differential equation

6) M(x, y) dx + N(x, y) dy = 0

the quantity

M(x, y) dx + N(x, y) dy

happens to be exactly the total differential of some function u(x, y) i.e. if there exists some function u(x, y) such that

then 6) is an exact differential equation and its primitive is u(x, y) = C.

Since ordinarily one cannot determine by inspection whether or not a given equation is exact, a test for exactness is necessary. That test is given by the following theorem.

Theorem 2. Let M(x, y), N(x, y), ∂M/∂y and ∂N/∂x be continuous functions of x and y. Then a necessary and sufficient condition that the differential equation

M(x, y) dx + N(x, y) dy = 0

be exact is

Proof

If the differential equation is exact the next step is to produce the function u(x, y) of which M(x, y) dx + N(x, y) dy is the total differential. Sometimes it can be determined from inspection. More often it cannot. The following example illustrates the usual method of solution.

Example. Test the following equation for exactness and find the solution if it is exact.

7) (3x²y - y)dx + (x³ - x + 2y)dy = 0

Solution.

M = 3x² - y N = x³ - x + 2y

∂M/∂y = 3x² - 1 ∂N/∂x = 3x² - 1

Since ∂M/∂y = ∂N/∂x, the equation is exact i.e. there is a function u(x, y) of which the left-hand side of 7) is exactly the total differential. To find this function we integrate ∂u/∂x = M = 3x² - y with respect to x, holding y constant. We obtain

8) u(x, y) = ∫M ∂x = ∫(3x² - y)∂x = x³y - yx + φ(y)

where φ(y) consists of terms that are free from x ( ∫M ∂x denotes integration with respect to x, holding y constant). In the same way, we integrate ∂u/∂y = N = x³ - x + 2y with respect to y, holding x constant. We obtain

9) u(x, y) = ∫N ∂y = ∫( x³ - x + 2y)∂y = x³y - xy + y² + ψ(x)

where ψ(x) consists of terms that are free from y ( i.e. terms containing x only or constants).

Comparing 8) and 9) we see that the general solution is

x³y - xy + y² = C

Example from Middlemiss. Differential and Integral Calculus. p. 443

There is a formula that can be used to obtain the solution. It is:

Formula for general solution of exact equation M dx + N dy = 0. The general solution is given by

∫ M ∂x + ∫f(y)dy = C

where f(y) is composed of all the terms in N which are free from x (i.e. all terms not containing x — terms containing y only or constants) and ∫M ∂x denotes integration with respect to x, keeping y constant.

Alternate formula:

∫N ∂y + ∫f(x)dx = C

where f(x) is composed of all the terms in M which are free from y.

The above formulas will give the correct result in the vast majority of cases but they are not infallible and in exceptional cases may give an incorrect result. Consequently the solution should always be checked by substituting it into the original equation.

There is another more complicated formula that also gives the general solution:

Formula for the general solution.

where ∫M ∂x denotes integration with respect to x, keeping y constant.

Derivation

IV Integrating factors. If a differential equation is not exact it may be possible to make it exact by multiplying it through by some function. For example, the equation

11) y dx + 2x dy = 0

is not exact since ∂M/∂y ≠ ∂N/∂x. However, multiplying it by y gives the exact equation

y²dx + 2xy dy = 0

in which the left-hand side is exactly the differential of xy². A function which, when multiplied into a differential equation, makes it exact is called an integrating factor. In this example y is an integrating factor for 11).

Theorem 3. Let the equation

12) M(x, y) dx + N(x, y) dy = 0

possess a solution f(x, y, C) = 0, where C is an arbitrary constant. Then if equation 12) is not exact, it can always be made exact by multiplying it through by some proper function of x and y i.e. there exists some integrating factor μ(x, y) that will make it exact. Moreover, if μ(x, y) is an integrating factor for 12) then a · μ(x, y) is also an integrating factor, where a is an arbitrary constant.

Thus if 12) is solvable, it is either exact or can be made exact by some integrating factor. There is however no general rule for finding that integrating factor. The theorem merely assures us that an exact version the equation exists.

The conditions for 12) to be solvable are not very severe. They are stated in Theorem 1 above.

There are many equations which are not integrable as they stand but which become integrable when multiplied by the right factor. There is no general method for finding the right factor but in many simple cases one can be found by inspection. The ability to do this depends largely upon recognition of certain common exact differentials and upon experience. One employs ingenuity and trial and error to, through manipulation, possibly transformation of variables, and some multiplying factor, transform the equation into one that can be integrated. For example, if one notices the groups of terms “x dy - y dx” or “x dy + y dx” in an equation he may be able to manipulate the equation into an equation that can be integrated using a multiplying factor that creates one of the following exact differentials:

Example. Solve the equation

(x² + y² + y)dx - x dy = 0

Solution. Let us write the equation as

(x² + y²)dx + y dx - x dy = 0

Remembering the formula

we decide to multiply the equation through by the factor 1/(x² + y²) to obtain

We then integrate to obtain the solution

Theorem 4. A differential equation of the form

can be reduced to the form

by means of the integrating factor

Its primitive is

Example. Solve the equation

(3x²y - xy) dx + (2x³y² + x³y⁴) dy = 0

Solution. We rewrite the equation as

y(3x² - x) dx + x³(2y² + y⁴) dy = 0

and multiply by the factor 1/yx³ which gives

Integrating, we get the primitive

Theorem 5.

a) A necessary and sufficient condition for μ(x, y) to be an integrating factor for the equation

13) M(x, y) dx + N(x, y) dy = 0

is that it satisfy the equation

b) If the quantity

is a function of x alone, i.e.

then

is an integrating factor for the equation

M dx + N dy = 0.

c) If the quantity

is a function of y alone, i.e.

then

is an integrating factor for the equation

M dx + N dy = 0.

Proof

Example. Solve y dx + (3 + 3x - y) dy = 0

Solution.

M = y N = 3 + 3x - y

∂M/∂y = 1 ∂N/∂x = 3

Now

is not a function of x alone, but

is a function of y alone. Hence an integrating factor is given by

If one multiplies the equation by y², one can confirm that it becomes exact and that its solution is

xy³ + y³ - y⁴/4 = C

V Linear equations of first order. A linear equation of first order is an equation of type

This equation has

as an integrating factor. The general solution is given by

Proof

Example. Solve

Solution.

P = 2/x

Multiplying both sides of the equation by this factor and integrating we have

x²y = x⁶ + C

Bernoulli’s equation. The equation

is known as Bernoulli’s equation. It can be transformed into a linear equation by the transformation

14) y ^{- n+1} = v

where v is a new variable.

Let us divide 13) by yⁿ to obtain the equivalent equation

If we now take the derivative of 14) with respect to x, we obtain

Substituting 14) and 16) into 15) yields

which is a linear equation in the variable v. We can now solve this equation by the method for linear equations.

The general solution for Bernoulli’s equation is

References

1. Ross R. Middlemiss. Differential and Integral Calculus. Chap. XXIX

2. James/James. Mathematics Dictionary.

3. Murray R. Spiegel. Applied Differential Equations.

4. James B. Scarborough. Differential Equations and Applications.

5. Frank Ayres. Differential Equations (Schaum).

6. Eshbach. Handbook of Engineering Fundamentals.

7. Earl Rainville. Elementary Differential Equations.