JUST THE MATHS SLIDES NUMBER 15.3 ORDINARY DIFFERENTIAL EQUATIONS - - PDF document

“JUST THE MATHS” SLIDES NUMBER 15.3 ORDINARY DIFFERENTIAL EQUATIONS 3 (First order equations (C)) by A.J.Hobson

15.3.1 Linear equations 15.3.2 Bernouilli’s equation

UNIT 15.3 - ORDINARY DIFFERENTIAL EQUATIONS 3 FIRST ORDER EQUATIONS (C) 15.3.1 LINEAR EQUATIONS For certain kinds of first order differential equation, it is possible to multiply the equation throughout by a suit- able factor which converts it into an exact differential equation. EXAMPLE dy dx + 1 xy = x2 may be multiplied throughout by x to give xdy dx + y = x3. It may now be written d dx(xy) = x3.

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This equation has general solution xy = x4 4 + C, where C is an arbitrary constant. Notes: (i) The factor, x which has multiplied both sides of the dif- ferential equation serves as an “integrating factor”, but such factors cannot always be found by inspection. (i) We shall now develop a formula for determining inte- grating factors for what are known as “linear differ- ential equations”. DEFINITION A differential equation of the form dy dx + P(x)y = Q(x) is said to be “linear”.

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RESULT (a) Given the linear differential equation dy dx + P(x)y = Q(x), the function e

P(x) dx

is always an integrating factor; (b) On multiplying the differential equation throughout by this factor, its left hand side becomes d dx

• y × e

P(x) dx

• .

Proof Suppose that the function R(x) is an integrating factor. Then, in the equation R(x)dy dx + R(x)P(x)y = R(x)Q(x), the left hand side must be the exact derivative of some function of x.

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We can make it the derivative of R(x)y provided we can arrange that R(x)P(x) = d dx[R(x)]. This requirement can be interpreted as a differential equa- tion in which the variables R(x) and x may be separated as follows:

• 1

R(x) dR(x) =

P(x) dx.

Hence, ln R(x) =

P(x) dx.

That is, R(x) = e

P(x) dx.

The solution of the linear differential equation is obtained by integrating the formula d dx[y × R(x)] = R(x)P(x).

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Note: There is no need to include an arbitrary constant, C, when P(x) is integrated; C would introduce a constant factor of eC in the above result, which would then cancel out on multiplying by R(x). EXAMPLES

• 1. Determine the general solution of the differential equa-

tion dy dx + 1 xy = x2. Solution An integrating factor is e

• 1

x dx = eln x = x.

On multiplying by the integrating factor, d dx[y × x] = x3. Hence, yx = x4 4 + C, where C is an arbitrary constant.

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• 2. Determine the general solution of the differential equa-

tion dy dx + 2xy = 2e−x2. Solution An integrating factor is e

2x dx = ex2.

Hence, d dx

• y × ex2

= 2, giving yex2 = 2x + C, where C is an arbitrary constant.

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15.3.2 BERNOUILLI’S EQUATION This type of differential equation has the form dy dx + P(x)y = Q(x)yn. It may be converted to a linear differential equation by making the substitution z = y1−n. Proof The differential equation may be rewritten as y−ndy dx + P(x)y1−n = Q(x). Also, dz dx = (1 − n)y−ndy dx. Hence, the differential equation becomes 1 1 − n dz dx + P(x)z = Q(x).

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That is, dz dx + (1 − n)P(x)z = (1 − n)Q(x), which is a linear differential equation. Note: It is better not to regard this as a standard formula, but to apply the method of obtaining it in the case of particular examples. EXAMPLES

• 1. Determine the general solution of the differential equa-

tion xy − dy dx = y3e−x2. Solution The differential equation may be rewritten −y−3dy dx + x.y−2 = e−x2. Substituting z = y−2, dz dx = −2y−3dy dx. Hence, 1 2 dz dx + xz = e−x2

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• r

dz dx + 2xz = 2e−x2. An integrating factor for this equation is e

2x dx = ex2.

Thus, d dx

• zex2

= 2, giving zex2 = 2x + C, where C is an arbitrary constant. Finally, replacing z by y−2, y2 = ex2 2x + C.

• 2. Determine the general solution of the differential equa-

tion dy dx + y x = xy2. Solution The differential equation may be rewritten y−2dy dx + 1 x.y−1 = x.

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Substituting z = y−1, dz dx = −y−2dy dx. Thus, −dz dx + 1 x.z = x

• r

dz dx − 1 x.z = −x. An integrating factor for this equation is e

(−1

x) dx = e− ln x = 1

x. Hence, d dx

  z × 1

x

   = −1,

giving z x = −x + C, where C is an arbitrary constant. The general solution of the given differential equation is therefore 1 xy = −x + C or y = 1 Cx − x2.

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