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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
(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
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
2x2 ln 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.
(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
(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
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
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
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.
∫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.
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
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.
is an integrating factor for the equation
M dx + N dy = 0.
c) If the quantity
is a function of y alone, i.e.
is an integrating factor for the equation
M dx + N dy = 0.
Example. Solve y dx + (3 + 3x - y) dy = 0
M = y N = 3 + 3x - y
∂M/∂y = 1 ∂N/∂x = 3
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
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
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.
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