Question

3.(x-y)dy _ (x + y) dr = 0

Answer 1

The given differential equation is,

( x−y )dy−( x+y )dx=0

Calculate the value of dy dx by rearranging the given differential equation,

dy dx = ( x+y ) ( x−y ) (1)

Consider the function F( x,y ) is equal to the dy dx and calculate F( λx,λy ),

F( x,y )= ( x+y ) ( x−y )

Replace x with λx and y with λy,

F( λx,λy )= ( λx+λy ) ( λx−λy ) = λ( x+y ) λ( x−y ) = ( x+y ) ( x−y ) = λ 0 F( x,y )

Because, F( x,y ) is a homogenous equation of the zero degree. So, it is clear that dy dx is a homogenous differential equation.

Calculate dy dx by substitute the value y=vx,

y=vx

Differentiate the above equation,

dy dx =x dv dx +v dx dx =x dv dx +v

Substitute the value of dy dx in equation (1),

x dv dx +v= ( x+y ) ( x−y ) x dv dx +v= ( x+vx ) ( x−vx ) x dv dx +v= 1+v 1−v x dv dx = 1+v 1−v −v

Further simplify the differential equation,

x dv dx = 1+v−v+ v 2 1−v x dv dx = 1+ v 2 1−v 1−v 1+ v 2 dv= 1 x dx

Integrate both sides,

∫ 1−v 1+ v 2 dv = ∫ 1 x dx ∫ 1−v 1+ v 2 dv =log| x |+C (2)

Consider,

I= ∫ 1−v 1+ v 2 dv = ∫ ( 1 1+ v 2 − v 1+ v 2 )dv = tan −1 v− ∫ v 1+ v 2 dv

Put, 1+ v 2 =t

Differentiate both the sides of the above equation,

2vdv=dt dt=2vdv

Substitute the values.

I= tan −1 v− ∫ v t dt 2v = tan −1 v− 1 2 log| t | = tan −1 v− 1 2 log| 1+ v 2 |+c

Substitute the above value in the equation (2),

tan −1 v− 1 2 log| 1+ v 2 |+c=log| x |+C tan −1 ( y x )− 1 2 log| 1+ ( y x ) 2 |=log| x |+C tan −1 y x = 1 2 [ log| 1+ ( y x ) 2 |+2log| x | ]+C tan −1 y x = 1 2 log[ x 2 + y 2 ]+C

Thus, the general solution of differential equation is tan −1 y x = 1 2 log[ x 2 + y 2 ]+C.