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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 = 13 v 2 v 3 3v x dv dx = 13 v 2 v 3 3v v = 13 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 dv3 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 1r | 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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