DIFFERENTIAL EQUATIONS

A Note If the homogeneous differential equation is in the form F( , ) x y dy = where, F ( x , y ) is homogenous function of degree zero, then we make substitution y = i.e., x = vy and we proceed further to find the general solution as discussed above by writing F( , ) x y h Example Show that the differential equation ( x – y ) dy dx = x + y is homogeneous and solve it. Solution The given differential equation can be expressed as Let F( x , y ) = Now F( λ x , λ y ) = ) ( , ) f x y λ = λ ⋅ λ Therefore, F( x , y ) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation. Alternatively , =  g R.H.S.

of differential equation ( ) is of the form g     and so it is a homogeneous function of degree zero. Therefore, equation ( ) is a homogeneous differential equation. To solve it we make the substitution y = vx ... ( ) Differentiating equation ( ) with respect to, x we get ...

( ) Substituting the value of y and dy dx in equation ( ) we get = x dx = x dx = Integrating both sides of equation ( ), we get = – log | x | + C (Why?) Replacing v by y x , we get log tan log ( tan 2C log ( tan which is the general solution of the differential equation ( ) Example Show that the differential equation is homogeneous and solve it. Solution The given differential equation can be written as  + It is a differential equation of the form F( , x y dx = Here F( x , y ) =  + Replacing x by λ x and y by λ y , we get F( λ x , λ y ) = [ cos ] [F( , )] x y λ = λ λ  Thus, F( x , y ) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation. To solve it we make the substitution y = vx Differentiating equation ( ) with respect to x , we get ...

( ) Substituting the value of y and dy dx in equation ( ), we get x dx = + − x dx = cos v cos v dv = dx Therefore cos v dv dx