Point 2. Integrable types of the first order differential equations (of de - 1)

1. Separated de-1 (de-1 with separated variables)

Def. 9. The first order differential equation of the form

, ( 7 )

where the variables x, y are separated, is called that in separated variables (or separated differential equation of the first order).

Theorem 3. The general solution of the equation (7) has the next form:

( 8 )

where mean some primitives of the functions correspondingly.

■a) Let a function be a solution of the equation (7) that is

.

Integrating the last identity with respect to x we get the equality (8), namely

.

b) Inversely, let a function satisfies the equality (8) that is

.

Differentiating this identity we obtain

,

and so the function is a solution of the equation (7).

Therefore every solution of the equation (7) satisfies the equality (8) and vice versa.

c) To prove possibility of choice a value of C to satisfy an initial condition

( 9 )

let’s rewrite both primitives in the equality (8) in the form of a definite integral,

.

By the initial condition (9) we get

.■

Note 3. Taking we’ll get . Therefore the solution of Cau-chy problem for the equation (7) with an initial condition (9) can be written in the next simplest form

. (10)

Note 4. Integration of a differential equation (7) is reduced to solving of more simple problem namely to evaluation of primitives. It is of no importance that at least one of these primitives can’t be expressed in terms of elementary functions.

Ex. 4. The differential equation is that in separated variables. Its general solution is given by the next formula

.

Both primitives do not express in terms of elementary functions.

2. Separable de-1 (de-1 with separable variables)

Def. 10. The first order differential equation is called that with separable va-riables (or separable differential equation of the first order) if it can be reduced to an equation in separated variables.

A differential equation

( 11 )

is that separable provided . It’s sufficient to represent as , multiply by and divide by both sides of the equation,

.

Therefore the general solution of the equation (11) is

. ( 12 )

To the same type of separable differential equations concerns the next one

( 13 )

if . We get a separated differential equation dividing both sides of the given equation by ,

,

and the general solution of the equation is given by the formula

. ( 14 )

Ex. 5. A differential equation has the form (13) ( ) and so it is separable one. Dividing both sides by we get

,

hence

.

Note 5. For the sake of more simplicity we can represent an arbitrary constant C in different forms.

Let’s take for example instead C in the preceding ex. 5, so

.

Putting finally , we get the general solution of the equation in the next more simple form

or

.

Ex. 6 (economical problem). Current inventory level of some company is 1680 items and it is producing with the rate of 900 items per month (i.p.m.). Demand is currently with the rate 800 i.p.m. and is dropping with the rate of 10 i.p.m. The company would like to reduce production with the rate n i.p.m. to sell all the production during the year. Find the value of n.

Let is the inventory level at a time moment t, and . Rate of its producing at a moment t is

.

It is known that

,

where is the rate of production, is the rate of selling. By conditions of the problem

,

or

.

We get the first order separable differential equation in with an initial condition

,

that is we must solve Cauchy problem. From the equation

,

by the initial condition , and

.

To sell all the production during the year we must have

,

whence it follows that .

Answer: its necessary to reduce the production with the rate of n = 50 i.p.m.

Ex. 7 (demographic problem). The rate of growth of population at arbitrary moment of time is proportional to the number of population at this moment (coefficient of proportionality equals k). Find the number of population at arbitrary moment t if it equals at the initial moment .

Let be the number of population at a moment t (it is obvious that ). The rate of growth of population at this moment

.

By the condition

.

Therefore,

,

and we have to solve Cauchy problem.

The differential equation of the problem is separable one,

,

.

Thus we’ve got exponential growth of population provided there are not opposed factors (decline in living standards, measures for limitation of birth rate etc.).

Ex. 8 (geometric problem). Find a curve through a point if a segment of its arbitrary tangent between the tangency point and the -axis is divided by the inter-section point of the segment and the -axis in the given Fig. 1 ratio 5 : 8, counting from the -axis (see fig. 1).

It follows from the conditions of the problem that an unknown curve can’t intersect the coordinate axes and therefore lies in the first quadrant.

Let is an unknown equation of a curve in question, an arbitrary point of the curve which is the tangency point, . Then

, (fig. 1).

By condition

We’ve Cauchy problem for a separable differential equation. Separating the variables we get

,

.

Using the initial condition we find the corresponding value of C and the solution of Cauchy problem,

.

Remark. There is the other way to get a differential equation of the problem proceeding from the equation of the tangent to a desired curve at its arbitrary point .

Let’s denote coordinates of an arbitrary point of the tangent by then the equation of the tangent will get the next form

.

Putting in this equation we’ll get

.

Making use of the conditions of the problem we obtain

hence

.