Question

Prove that is the general solution of differential equation , where c is a parameter.

Answer 1

Given, differential equation is ( x 3 −3x y 2 )dx=( y 3 −3 x 2 y )dy.

Simplify the above equation.

( x 3 −3x y 2 )dx=( y 3 −3 x 2 y )dy dy dx = x 3 −3x y 2 y 3 −3 x 2 y (1)

The above equation is homogeneous equation. Substitute y=vx and further simplify.

y=vx dy dx =v+x dv dx

Substitute the value of y and dy dx in the equation (1) we get,

v+x dv dx =[ x 3 −3x ( vx ) 2 ( vx ) 3 −3 x 2 ( vx ) ] v+x dv dx = 1−3 v 2 v 3 −3v x dv dx = 1−3 v 2 v 3 −3v −v = 1−3 v 2 −v( v 3 −3v ) v 3 −3v

Further simplify.

x dv dx = 1− v 4 v 3 −3v ( v 3 −3v 1− v 4 )dv= dx x

Integrate both sides of above equation.

∫ ( v 3 −3v 1− v 4 )dv =logx+log C ′ (2)

Consider the Left Hand Side of the above equation.

L.H.S.= ∫ ( v 3 −3v 1− v 4 )dv = ∫ v 3 1− v 4 dv−3 ∫ vdv 1− v 4 (3)

Let, I 1 = ∫ v 3 1− v 4 dv and I 2 = ∫ vdv 1− v 4 .

Solve for I 1 .

1− v 4 =t d dv ( 1− v 4 )= dt dv −4 v 3 = dt dv v 3 dv=− dt 4

Now substitute the above values in I 1 .

I 1 = ∫ −dt 4t =− 1 4 logt I 1 =− 1 4 log( 1− v 4 )

Now, solve for I 2 .

I 2 = ∫ vdv 1− ( v 2 ) 2

Let, v 2 =r.

d dv ( v 2 )= dr dv 2v= dr dv vdv= dr 2

Substitute the above value in I 2 .

I 2 = 1 2 ∫ dr 1− r 2 = 1 2×2 log| 1+r 1−r | I 2 = 1 4 log| 1+ v 2 1− v 2 |

Substitute the above value in equation (3) we get,

L.H.S.= I 1 −3 I 2 =− 1 4 log( 1− v 4 )− 3 4 log| 1+ v 2 1− v 2 |

Substitute the above value in the equation (2).

− 1 4 log( 1− v 4 )− 3 4 log| 1+ v 2 1− v 2 |=logx+log C ′ − 1 4 log[ ( 1− v 4 )× ( 1+ v 2 1− v 2 ) 3 ]=log C ′ x ( 1+ v 2 ) 4 ( 1− v 2 ) 2 = ( C ′ x ) −4 ( 1+ y 2 x 2 ) 4 ( 1− y 2 x 2 ) = 1 C ′ 4 x 4

Further simplify.

( x 2 + y 2 ) 4 x 4 ( x 2 − y 2 ) 2 = 1 C ′ 4 x 4 { ( x 2 − y 2 ) 2 }= C ′ 4 ( x 2 + y 2 ) 4 { ( x 2 − y 2 ) }= C ′ 2 ( x 2 + y 2 ) 2 x 2 − y 2 =C ( x 2 + y 2 ) 2

Hence, the given result proved.