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Question

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

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Solution

## 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.

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