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 d y d x , λ may be removed from the equation by cancelling a common factor of λ n , for some integer n . Note: Some examples of homogeneous equations would be ( x + y )d y d x + (4 x − y ) = 0 and 2 xy d y d x + ( x 2 + y 2 ) = 0 . From the first of these, a factor of λ could be cancelled. From the second, a factor of λ 2 could be cancelled. 1

15.2.2 THE STANDARD METHOD We make the substitution y = vx , giving d x = v + x d v d y d x. 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 x d y d x = x + 2 y, given that y = 6 when x = 6. Solution If y = vx , then d y d x = v + x d v d x , so that  v + x d y    = x + 2 vx. x   d x 2

That is, v + x d v d x = 1 + 2 v or x d v d x = 1 + v. On separating the variables, � 1 1 � 1 + v d v = x d x, 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 or y = Ax 2 − x. Finally, if y = 6 when x = 1, we have 6 = A − 1 and hence A = 7, giving y = 7 x 2 − x . 3

2. Determine the general solution of the differential equa- tion ( x + y )d y d x + (4 x − y ) = 0 . Solution If y = vx , then d y d x = v + x d v d x , so that  v + x d v    + (4 x − vx ) = 0 . ( x + vx )   d x That is,  v + x d v    + (4 − v ) = 0 (1 + v )   d x or v + x d v d x = v − 4 v + 1 . On further rearrangement, v + 1 − v = − 4 − v 2 x d v d x = v − 4 v + 1 . On separating the variables, � 1 v + 1 � 4 + v 2 d v = − x d x or � 1 1 2 v 2    d v = − � 4 + v 2 + x d x.    2 4 + v 2 4

Hence, 1  ln(4 + v 2 ) + tan − 1 v    = − ln x + C, 2 2 where C is an arbitrary constant. Substituting back v = y x ,  4 + y 2 1     y     + tan − 1  ln  = − ln x + C.          2 x 2 2 x 3. Determine the general solution of the differential equa- tion 2 xy d y d x + ( x 2 + y 2 ) = 0 . Solution If y = vx , then d y d x = v + x d v d x , so that  v + x d v    + ( x 2 + v 2 x 2 ) = 0 . 2 vx 2   d x That is,  v + x d v    + (1 + v 2 ) = 0 2 v   d x or 2 vx d v d x = − (1 + 3 v 2 ) . 5

On separating the variables, � 1 2 v � 1 + 3 v 2 d x = − x d x, which gives 1 3 ln(1 + 3 v 2 ) = − ln x + ln A, where A is an arbitrary constant. Hence, 3 = A 1 (1 + 3 v 2 ) x. On substituting back v = y x , 1  x 2 + 3 y 2   3 = Ax.     x 2  This can be written x 2 + 3 y 2 = Bx 5 , where B = A 3 . 6