10.6.3 Homogeneous Form or Homogeneous Differential Equation

. . Homogeneous Form or Homogeneous Differential Equation Definition . : (Homogeneous Function of degree n ) A function f x y ( , ) is said to be a homogeneous function of degree n in the variables x and y if, f tx ty t f x y ( , for some n ∈  for all suitably restricted x y and t .

This is known as Euler’s homogeneity . For instance, (i) f x y ( , ) = is a homogeneous function in x and y , of degree two. (ii) But f x y x e y + ( is not a homogeneous function. If f x y ( , ) is a homogeneous function of degree zero, then there exists a function g such that f x y ( , ) is always expressed in the form g   or g   .

Definition . : (Homogeneous Differential Equation) An ordinary differential equation is said to be in homogeneous form , if the differential equation is written as dy g   . Caution The word “homogeneous” used in Definition . is different from in Definition .

. Remark (i) The differential equation M x y dx N x y dy = [in differential form] is said to be homogeneous if M and N are homogeneous functions of the same degree . (ii) The above equation is also written as dy f x y ( , ) [in derivative form] where f x y M x y N x y ( , ) / = − is clearly homogeneous of degree . For instance ( ) consider the differential equation x xy dy .

The given equation is rewritten as  − y x / . Thus, the given equation is expressed as y x g  −  = / . Hence, xy dy is a homogeneous differential equation. Ordinary Differential Equations ( ) However, the differential equation dy is not homogeneous.

(verify!) To find the solution of a homogeneous differential equation dy g   , consider the substitution v . Then, y xv and dy v x dv .Thus, the given differential equation becomes x dv f v v which is solved using variables separable method. This leads to the following result. Theorem .

If M x y dx N x y dy = is a homogeneous differential equation, then the change of variable vx , transforms into a separable equation in the variables v and x . Example . Solve x xydy We know that the given equation is homogeneous. Now, we rewrite the given equation as dy dx = Taking y vx , we have v x dv = v v or x dv v v Separating the variables, we obtain vdv v − = dx x .

On integration, we get log v − = log Hence v − = Cx , where C is an arbitrary constant. Now, replace v by y x to get y = Cx . Thus, we have y = Cx . Hence, y = ± Cx (or) y kx gives the general solution.

Example . Solve y xdy The given differential equation is homogeneous (verify!). Now, we rewrite the given equation in differential form dy Since the initial value of x is , we consider x > and take x . We have dy +  Let y vx .

Then, v x dv v v , which becomes x dv v . By separating variables, we have dv v Upon integration, we get log v v or v v xC + = Now, we replace v by y x , we get y Cx + = (or) y Cx gives the general solution of the given differential equation. To determine the value of C , we use the condition that y = when x = . So, we get C = .

Thus y is the particular solution of the given differential equation. Example . Solve y dx x dy The given equation can be written as dx = This is a homogeneous equation. Let y vx .

Then we have v x dv = v v . Thus, x dv v v v or ) + v v dv x or − ) +       v v v v dv x . Integrating both sides, we get − ) = tan v v v or log tan v v v ) = − or log tan v v v ) = − or log tan v v v ) = − Now replacing v by y x , we get, log tan k  = , where k = − 2log gives the required solution. Ordinary Differential Equations Example .

Solve y x dy xy dy The given equation is rewritten as dy dx = This is a homogeneous differential equation. Put y vx . Then, we have x dv dx = v v − . By separating the variables, v v dv − = dx x .

Integrating, we obtain v v − log = log or v vxC = log Replacing v by y x , we get, y Cy = log or Cy e y x / or y ke y x / (how!) which is the required solution. Example . Solve y dy x y x y / / The given equation can be written as dx g x y x y / / …( ) The appearance of x y in equation ( ), suggests that the appropriate substitution is x vy Put x vy . Then, we have y dv v v v = − By separating the variables, we have = − v e dv v v On integration, we obtain e v v + = − or log ye vy v + or ye vy v + = ± Replace v by x y to get, ye k x y / + , where k = ± ,which gives the required solution.