Consider the following functions in x and y

`color{red} {F_1 (x, y) = y^2 + 2xy, F_2 (x, y) = 2x – 3y}`,

`color{red} {F_3 (x,y) = cos (y/x) , F_4 (x, y) = sin x + cos y}`

If we replace x and y by `λx` and `λy` respectively in the above functions, for any nonzero constant `λ,` we get

`F_1 (λx, λy) = λ^2 (y^2 + 2xy) = λ^2 F_1 (x, y)`

`F_2 (λx, λy) = λ (2x – 3y) = λ F_2 (x, y)`

`F_3 (λx, λy) = cos ((lamday)/(lamdax)) = cos (y/x) = lamda^0 F_3 (x,y)`

`F_4 ( λx, λy) = sin λx + cos λy ≠ λ^n F_4 (x, y)`, for any `n ∈ N`

`\color{green} ✍️` Here, we observe that the functions `F_1, F_2, F_3` can be written in the form `F(λx, λy) = λ^n F (x, y)` but `F_4` can not be written in this form. This leads to the following definition:

`color{blue}{"A function F(x, y) is said to be homogeneous function of degree n if" F(λx, λy) = λ^n F(x, y) "for any nonzero constant λ."}`

`=>` We note that in the above examples, `F_1, F_2, F_3` are homogeneous functions of degree 2, 1, 0 respectively but `F_4` is not a homogeneous function.

We also observe that

`F_1(x,y) = x^2 (y^2/x^2 + (2y)/x) = x^2 h_1 (y/x)`

or `F_1(x,y) = y^2 ( 1 + (2x)/y) = y^2 h_2 (x/y)`

`F_2 (x,y) = x^1 (2 - (3y)/x ) = x^1h_3 (x/y)`

or `F_2 (x,y) = y^1(2 x/y -3) = y^1h_4(x/y)`

`F_3 (x,y) =x^0 cos (y/x) =x^0 h_5 (y/x)`

`F_4 (x,y) != x^n h_6 (y/x)`, for any `n in N`

or ` F_4 (x,y) != y^n h_1 (x/y)`, for any `n in N`

`=>` Therefore, a function `F (x, y)` is a homogeneous function of degree `n` if

`color{blue}{F(x,y) = x^n g(y/x) \ \ "or" \ \ y^nh(x/y)}`

`\color{green} ✍️` `(dy)/(dx) = F (x, y)` is said to be homogenous if `F(x, y)` is a homogenous function of degree zero.


`color{red}{"Simplification Techniques"}`

`=>` To solve a homogeneous differential equation of the type

`color{red} {(dy)/(dx) =F(x,y) = g (y/x)}` ........................(1)

`=>` We make the substitution `color{red} {y = v . x}` ... (2)

`=>` Differentiating equation (2) with respect to `x,` we get

`color{red} {(dy)/(dx) = v + x (dv)/(dx)}` ...............................(3)

`=>` Substituting the value of `(dy)/(dx)` from equation (3) in equation (1), we get

`v + x (dv)/(dx) = g(v)`

or `color{red} {x(dv)/(dx) = g(v) - v}` ..............................(4)

`=>` Separating the variables in equation (4), we get

`color{red} {(dv)/(g(v) -v) = (dx)/x}` .............................(5)

Integrating both sides of equation (5), we get

`color{red} {int(dv)/(g(v) -v) = int 1/x dx +C}` ....................(6)

Equation (6) gives general solution (primitive) of the differential equation (1) when we replace `color{green}v` by `color{orange}{y/x}`


`"Key Concept :"`
If the homogeneous differential equation is in the form `(dx)/(dy) = F(x,y)` where, F (x, y) is homogenous function of degree zero, then we make substitution `x/y =v` i.e., `x = vy` and we proceed further to find the general solution as discussed above by writing `(dx)/(dy) = F(x,y) = h(x/y)`