Consider the differential equation \(2xy\displaystyle\frac{dy}{dx} = x^2 + 3y^2\), \(x \gt 0\), \(y \gt 0\).
(A) It is a homogeneous differential equation.
(B) The substitution \(y = vx\) reduces it to \(x\displaystyle\frac{dv}{dx} = \frac{1 + 3v^2}{2v}\).
(C) Its general solution is \(x^2 + y^2 = Cx^3\), where C is an arbitrary constant.
(D) The solution curve through the point (1, 1) also passes through the point \((2, 2\sqrt{3})\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C) and (D) only
The equation is
\( \displaystyle \frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy} \)
The right side is homogeneous of degree zero, so the equation is homogeneous. With \(y = vx\):
\( \displaystyle v + x\frac{dv}{dx} = \frac{1 + 3v^2}{2v} \)
\( \displaystyle x\frac{dv}{dx} = \frac{1 + 3v^2 - 2v^2}{2v} = \frac{1 + v^2}{2v} \)
Separating and integrating:
\( \displaystyle \int \frac{2v}{1 + v^2}\,dv = \int \frac{dx}{x} \)
\( \log(1 + v^2) = \log x + \log C \)
\( \displaystyle 1 + \frac{y^2}{x^2} = Cx \)
\( x^2 + y^2 = Cx^3 \)
Through (1, 1): \(2 = C\), so the curve is \(x^2 + y^2 = 2x^3\). At \(x = 2\):
\( y^2 = 16 - 4 = 12 \Rightarrow y = 2\sqrt{3} \)
So the equation is homogeneous, it reduces to \(x\frac{dv}{dx} = \frac{1 + v^2}{2v}\), its general solution is \(x^2 + y^2 = Cx^3\), and the curve through (1, 1) gives \(y = 2\sqrt{3}\) at \(x = 2\).
Checking the statements
(A) It is a homogeneous differential equation. — Correct. \(\frac{x^2 + 3y^2}{2xy}\) is homogeneous of degree zero.
(B) The substitution \(y = vx\) reduces it to \(x\frac{dv}{dx} = \frac{1 + 3v^2}{2v}\). — Incorrect. The v on the left must be subtracted: \(x\frac{dv}{dx} = \frac{1 + v^2}{2v}\).
(C) Its general solution is \(x^2 + y^2 = Cx^3\). — Correct. \(\log(1 + v^2) = \log Cx\) gives \(x^2 + y^2 = Cx^3\).
(D) The solution curve through (1, 1) also passes through \((2, 2\sqrt{3})\). — Correct. \(x^2 + y^2 = 2x^3\): \(4 + 12 = 16\).