Question

Please solve (y³-3x²y)dy=(x³-3xy²)dx

Answer 1

(y³ - 3x²y)dy = (x³ - 3xy²)dx

y³.dy - 3x²y.dy = x³.dx - 3xy².dx

(y³dy - x³dx ) = 3x²y.dy - 3xy².dx
divided by x³ both sides,
(y/x)³.dy - dx = 3(y/x).dy - 3(y/x)²dx
dy{(y/x)³ -3(y/x) } = dx{1 - 3(y/x)² }


y/x = P ( Let )
then,
dy/dx = ( 1 - 3P²)/(P³ -3P)

now,
y = Px
differentiate with respect to x
dy/dx = P + xdP/dx

P + x.dP/dx = (1 - 3P²)/(P³- 3P) - P
x.dP/dx = { (1 - 3P²) - P(P³ - 3P) }/(P³ - 3P)
x.dP/dx = ( 1 - P⁴)/(P³ - 3P)
(P³ - 3P)dP/(1 - P⁴) = dx/x
integrate both sides,
-1/4 { -4P³.dP/(1 - P⁴)} -3P.dP/( 1 - P⁴) = lnx + C
-1/4ln( 1 - P⁴) - R = lnx + C

Where R = 3P.dP/( 1 - P⁴)
Let P² = z
2P.dP = dz
P.dP = dz/2
R = 3/2dz/(1 - z²) = 3/4ln(1 - z)/(1 + z)

now , R = 3/4 ln(1 - P²)/(1 + P²)

hence,

-1/4ln( 1 - P⁴) -3/4ln(1 - P²)/( 1+ P²) = lnx + C
put P = y/x

-1/4ln(1 - y⁴/x⁴) -3/4ln( x² - y²)/(x² + y²) = lnx + C

Answer 2

-1/4ln(1-y⁴/x⁴) -3/4ln( x²-y²)/(x²+y²)=lnx+C

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