Question

4. (x2 + 2.xy dy = 0

Answer 1

The given differential equation is,

( x 2 − y 2 )dx+2xydy=0

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

dy dx = ( y 2 − x 2 ) 2xy (1)

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

F( x,y )= ( y 2 − x 2 ) 2xy

Replace x by λx and y by λy,

F( λx,λy )= ( ( λy ) 2 − ( λx ) 2 ) 2( λx )( λy ) = λ 2 ( y 2 − x 2 ) λ 2 ( 2xy ) = ( y 2 − x 2 ) ( 2xy ) = λ 0 F( x,y )

Because, F( x,y ) is a homogenous equation with zero degree. So, 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= ( ( vx ) 2 − x 2 ) ( 2x( vx ) ) x dv dx +v= ( v 2 −1 ) 2v x dv dx = ( v 2 −1 ) 2v −v x dv dx = −( v 2 +1 ) 2v

Further, simplify the differential equation,

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

Integrate both sides,

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

Consider,

I= ∫ 2v 1+ v 2 dv = ∫ 2v 1+ v 2 dv

Put, 1+ v 2 =t

Differentiate both the sides of the above equation,

2vdv=dt dt=2vdv

Substitute the above value in I,

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

Substitute the value of v,

I=log| 1+ ( y x ) 2 |+c

Substitute the above value in the equation (2),

log| 1+ ( y x ) 2 |=−log| x |+C log| 1+ ( y x ) 2 |+log| x |=C log| ( 1+ ( y x ) 2 )×x |=C log| x 2 + y 2 x |=logC

Further, simplify the above equation,

x 2 + y 2 =Cx

Thus, the general solution of differential equation is x 2 + y 2 =Cx.