VIEWS: 29 PAGES: 4 CATEGORY: Technology POSTED ON: 7/9/2010 Public Domain
Homogeneous Equations A homogeneous equation can be transformed into a separable equation by a change of variables. Definition: An equation in differential form M(x,y) dx + N(x,y) dy = 0 is said to be homogeneous, if when written in derivative form dy ! y$ = f(x, y) = g # & dx " x% ! y$ there exists a function g such that f(x,y) = g # & . " x% Example: (x2 – 3y2) dx – xy dy = 0 is homogeneous since !1 dy x 2 ! 3y 2 x y " y% y " y% = = ! 3 = $ ' ! 3 = g$ ' . dx xy y x # x& x # x& We can use another approach to define a homogeneous equation. Definition: A function F(x,y) of the variables x and y is called homogeneous of degree n if for any parameter t F(tx, ty) = tn F(x,y) Example: Given F(x,y) = x3 – 4x2y + y3, it is a homogeneous function of degree 3 since F(tx,ty) = (tx)3 – 4(tx)2(ty) + (ty)3 = t3(x3 – 4x2y + y3). Theorem: The O.D.E. in differential form M(x,y) dx + N(x,y) = 0 is a homogeneous O.D.E. if M(x,y) and N(x,y) are homogeneous functions of the same degree. Proof: Assume M(x,y) and N(x,y) are homogeneous functions of degree n, then M(tx,ty) = tnM(x,y) and N(x,y) = tnN(x,y) Assume that the parameter t = 1/x, then n n ! y$ ! 1 1 $ ! 1$ ! y$ ! 1 1 $ ! 1$ M # 1, & = M # x, y & = # & M(x, y)!and!N # 1, & = N # x, y & = # & N(x, y) " x% " x x % " x% " x% " x x % " x% then, ( x )n M " 1, % " y% y $ # x& ' M $ 1, ' # x& dy M(x, y) " y% =! =! =! = g$ ' # x& ( x )n N " 1, % " y% dx N(x, y) y $ # x& ' N $ 1, ' # x& this shows that the equation is homogeneous in the sense of the first definition. Theorem: Given a homogeneous O.D.E., the change of variable y = vx transforms the equation into a separable equation in the variables v and x. Proof: dy ! y$ Given the homogeneous O.D.E. = g# & , dx " x% y Let y = vx or v = (notice that v depends on x) x dy dv then = v+ x dx dx and the equation is transformed into dv v+ x = g(v)!or![ v ! g(v)] dx + xdv = 0 , it is a separable equation. dx If we separate the variables, we get dv dx + =0 v ! g(v) x integrating, we get dv dx " v ! g(v) + " x = c !or!!G(v) + ln x = c = ln m since we can name the arbitrary constant any way we want. Example: Solve the equations 1) (x2 – 3y2) dx + 2xy dy = 0 M(x,y) = x2 – 3y2 and N(x,y) = 2xy are homogeneous functions of degree 2. Let’s express the equation in derivative form: dy x 3y =! + dx 2y 2x dy dv Take the transformation y = vx and = v+ x dx dx then, dv 1 3v v+ x =! + dx 2v 2 or dv 1 3v !2v2 ! 1 + 3v2 v2 ! 1 x = !v ! + = = dx 2v 2 2v 2v separating variables 2v dx dv = v !1 2 x integrating 2v dx " v2 ! 1 dv = " x ln v2 ! 1 = ln x + ln c v2 ! 1 = xc replacing v = y/x, y2 2 ! 1 = xc !or! y 2 ! x 2 = xc x 2 x 2) Solve the I.V.P. (x2 –xy + y2)dx – xy dy = 0 y(1) = 0 the equation in derivative form is dy x 2 ! xy ! y 2 x y " y% = = ! 1+ = g$ ' dx xy y x # x& it is homogeneous. Take the transformation y = vx, and replace dv 1 dv 1 1! v v+ x = ! 1 + v!!or!x = !1= dx v dx v v separate the variables and integrate v dx dv = 1! v x dx v " x + " v ! 1 dv = c dx v !1+1 " x + " v ! 1 dv = c dx v!1 dv " x + " v ! 1 dv!+ " v ! 1 = c ln x + v + ln v ! 1 = ln m v = ! ln ( v ! 1 x m ) e !v = v ! 1 x m y y y s = ! 1 x e x !or! s = y ! x e x x using the initial conditions y = 0 when x = 1, |s|= |0 – 1| e0 = 1 then s = ±1, but |y – x| > 0 and ey/x > 0, then s = 1 The solution I.V.P. is: 1 = |y – x| ey/x 3) (x ey/x – y)dx + x dy = 0 the equation in derivative form is: y dy y = !e x + dx x take the transformation y = vx dv dv v+ x = !e v + v!!or!!x = !e v dx dx separate variables dx !e vdv = x dx " !e dv = " x !v e !v = ln x + ln c = ln ( x c ) y ! e x = ln ( xc )