The curve satisfying the differential equation ydx − (x + 3y²)dy = 0 and passing through the point (1, 1), also passes through the point. Options: (14, 12), (13, -13), (-13, 13), (14, -12)

The given differential equation is ydx - (x + 3y²)dy = 0.
This can be rewritten as ydx = (x + 3y²)dy.
Dividing by y², we get:
(ydx)/y² = (x/y² + 3)dy
(dx/y) = (x/y² + 3)dy
(dx/y) - 3dy = (x/y²)dy
This is a linear differential equation of the form:
dX/dy + P(y)X = Q(y)
where X = x, P(y) = -1/y², Q(y) = 3
The integrating factor is given by:
IF = e^(∫P(y)dy) = e^(∫-1/y²dy) = e^(1/y)
Multiplying the equation by the integrating factor, we get:
e^(1/y)(dx/y) - 3e^(1/y)dy = e^(1/y)(x/y²)dy
This simplifies to:
d/dy[xe^(1/y)] = 3e^(1/y)dy
Integrating both sides, we get:
xe^(1/y) = ∫3e^(1/y)dy
Let u = 1/y, then du = -dy/y² , dy = -y²du = -du/u²
xe^(1/y) = ∫3e^u(-du/u²) = -3∫(e^u/u²)du
This integral is not easily solvable.
Let's try another approach.
The given differential equation is ydx - (x + 3y²)dy = 0.
This is a homogeneous differential equation.
Let x = vy, then dx = vdy + ydv
Substituting into the equation, we get:
y(vdy + ydv) - (vy + 3y²)dy = 0
vydy + y²dv - vydy - 3y²dy = 0
y²dv - 3y²dy = 0
dv = 3dy
Integrating both sides, we have:
v = 3y + c
Replacing v = x/y, we get:
x/y = 3y + c
x = 3y² + cy
Since the curve passes through (1, 1), we have:
1 = 3(1)² + c(1)
c = 1 - 3 = -2
Therefore, the equation of the curve is x = 3y² - 2y.
Let's check the options:
For (14, 12): 14 = 3(12)² - 2(12) = 432 - 24 = 408 (False)
For (13, -13): 13 = 3(-13)² - 2(-13) = 507 + 26 = 533 (False)
For (-13, 13): -13 = 3(13)² - 2(13) = 507 - 26 = 481 (False)
For (14, -12): This is not possible since y must be positive.
There must be a mistake in the question or options provided.