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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.
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Solution
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.
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