JUST THE MATHS SLIDES NUMBER 15.2 ORDINARY DIFFERENTIAL EQUATIONS - - PDF document

“JUST THE MATHS” SLIDES NUMBER 15.2 ORDINARY DIFFERENTIAL EQUATIONS 2 (First order equations (B)) by A.J.Hobson

15.2.1 Homogeneous equations 15.2.2 The standard method

UNIT 15.2 - ORDINARY DIFFERENTIAL EQUATIONS 2 FIRST ORDER EQUATIONS (B) 15.2.1 HOMOGENEOUS EQUATIONS A differential equation of the first order is said to be “ho- mogeneous” if, on replacing x by λx and y by λy in all the parts of the equation except dy

dx, λ may be removed

from the equation by cancelling a common factor of λn, for some integer n. Note: Some examples of homogeneous equations would be (x + y)dy dx + (4x − y) = 0 and 2xydy dx + (x2 + y2) = 0. From the first of these, a factor of λ could be cancelled. From the second, a factor of λ2 could be cancelled.

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15.2.2 THE STANDARD METHOD We make the substitution y = vx , giving dy dx = v + xdv dx. This always converts a homogeneous differential equation into one in which the variables can be separated. The method will be illustrated by examples: EXAMPLES

• 1. Solve the differential equation

xdy dx = x + 2y, given that y = 6 when x = 6. Solution If y = vx, then dy

dx = v + xdv dx, so that

x

  v + xdy

dx

   = x + 2vx. 2

That is, v + xdv dx = 1 + 2v

• r

xdv dx = 1 + v. On separating the variables,

• 1

1 + v dv =

1

x dx, giving ln(1 + v) = ln x + ln A, where A is an arbitrary constant. An alternative form of this solution, without logarithms, is Ax = 1 + v. Substituting back v = y

x,

Ax = 1 + y x

• r

y = Ax2 − x. Finally, if y = 6 when x = 1, we have 6 = A − 1 and hence A = 7, giving y = 7x2 − x.

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

tion (x + y)dy dx + (4x − y) = 0. Solution If y = vx, then dy

dx = v + xdv dx, so that

(x + vx)

  v + xdv

dx

   + (4x − vx) = 0.

That is, (1 + v)

  v + xdv

dx

   + (4 − v) = 0

• r

v + xdv dx = v − 4 v + 1. On further rearrangement, xdv dx = v − 4 v + 1 − v = −4 − v2 v + 1 . On separating the variables,

• v + 1

4 + v2 dv = −

1

x dx

• r

1 2

• 

 

2v 4 + v2 + 2 4 + v2

   dv = − 1

x dx.

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Hence, 1 2

 ln(4 + v2) + tan−1v

2

  = − ln x + C,

where C is an arbitrary constant. Substituting back v = y

x,

1 2

   ln    4 + y2

x2

    + tan−1   y

2x

      = − ln x + C.

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

tion 2xydy dx + (x2 + y2) = 0. Solution If y = vx, then dy

dx = v + xdv dx, so that

2vx2

  v + xdv

dx

   + (x2 + v2x2) = 0.

That is, 2v

  v + xdv

dx

   + (1 + v2) = 0

• r

2vxdv dx = −(1 + 3v2).

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On separating the variables,

• 2v

1 + 3v2 dx = −

1

x dx, which gives 1 3 ln(1 + 3v2) = − ln x + ln A, where A is an arbitrary constant. Hence, (1 + 3v2)

1 3 = A

x. On substituting back v = y

x,

   x2 + 3y2

x2

   

1 3

= Ax. This can be written x2 + 3y2 = Bx5, where B = A3.

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