Question

Prove that x² – y² = c (x² + y²)² is the general solution of differential equation  (x³ – 3x y²) dx = (y³ – 3x²y) dy, where c is a parameter.

Answer 1

We have, `dy/dx = (x^3 - 3xy^2)/(y^3 - 3x^2 y)`          ....(1)

Put y = vx

⇒ `dy/dx = v + x (dv)/dx`

∴ (1) become:

`v + x  (dv)/dx = (x^3 - 3x (v^2 x^2))/(v^3 x^3 - 3x^2 vx)`

`= (1 - 3v^2)/(v^3 - 3v)`

⇒ `x (dv)/dx = (1 - 3v^2)/(v^3 - 3v) - v`

`= (1 - 3v^2 - v^4 + 3v^2)/ (v^3 - 3v)`

`= (1 - v^4)/(v^3 - 3v)`

⇒ `(v^3 - 3v)/(1 - v^4)  dx = dx/x`

Integrating, `int (v^3 - 3v)/ (1 - v^4) dv = int dx/x + `constant    ....(2)

Now,

`I = int (v^3 - 3v)/ (1 - v^4)  dv`

`= int v^3/ (1 - v^4)  dv - 3 int v/ (1 - v^4)  dv`                 ....(3)

I = I₁ - 3I₂                             ....(4 )

Where `I = int v^3/(1 - v^4)  dv`

Put 1 - v⁴ = t

⇒ -4v³ dv = dt

⇒ `v^3 dv = -dt/4`

∴ `I_1 = int (-1/4  dt)/t`

`= 1/4 int 1/t dt = -1/4 log |t| + C_1`

`= -1/4 log |1 - v^4| + C_1`

And `I_2 - int v/ (1 - v^4)  dv`

Put v² = T

⇒ 2v = dT

⇒ `vdv = (dT)/2`

∴ `I_2 = int (1/2 dT)/ (1 - T^2)`

`= 1/2 int (dT)/(1^2 - T^2)`

`= 1/(2(2)) log |(1 + T)/(1 - T)| + C_2`

`= 1/4 log |(1 + v^2)/ (1 - v^2) + C_2|`

∴ From (4), we get

`I = 1/4  log |1 - v^4|  -3/4  log |(1 +v^2)/(1 - v^2)| + C_1 + C_2`

From (2), we have

`- 1/4 log |1 - v^4| - 3/4  log |(1 + v^2)/ (1 - v^2)|= log |x| + log |C'|`

⇒ `-1/4 [log |1 - v^4| + 3 log |(1 + v^2)/(1 - v^2)|] = log |C'  x|`

⇒ `-1/4 [log |(1 - v^2) (1 + v^2) (1 + v^2)^3/(1 - v^2)^3|] = log |C'  x|`

⇒ `-1/4 [log |(1 + v^2)^4/(1 - v^2)^2|] = log |C'  x |`

⇒ `log | sqrt (1 - v^2)/ (1 + v^2)| = log |C'  x|`

⇒ `sqrt (1 - v^2)/ (1 + v^2) = C'  x`

⇒ `sqrt (1 - y^2/x^2)/(1 + y^2/x^2) = C'  x`

⇒ `sqrt (x^2 y^2) = C' (x^2 + y^2)`

Squaring on the both sides, we get

`x^2 - y^2 = C (x^2 + y^2)^2`  Where C'² = C

Hence, x² - y² = C (x² + y²)² is the general solution.