Question

Let $y=y(x)$ be the solution of the differential equation $\left(3 y^2-5 x^2\right) y d x+2 x\left(x^2-y^2\right) d y=0$ such that $y(1)=1$ Then $\left|(y(2))^3-12 y(2)\right|$ is equal to :

• A. 64
• B. $32 \sqrt{2}$
• C. 32
• D. $16 \sqrt{2}$

Hint:

To solve such first-order differential equations, consider substitution or separation of variables. Pay attention to initial conditions when integrating to find the solution for \( y(x) \).

Answer 1

The Correct Option is B

We are given the differential equation \[ (3y^2 - 5x^2) y \, dx + 2x(x^2 - y^2) \, dy = 0. \] Step 1: Rearrange the equation \[ (3y^2 - 5x^2) y \, dx = -2x(x^2 - y^2) \, dy. \] Now divide both sides by \( y(x^2 - y^2) \), and separate variables: \[ \frac{(3y^2 - 5x^2)}{y(x^2 - y^2)} \, dx = -2 \, dy. \] Step 2: Integrate both sides. The integral on the left-hand side involves separating the terms: \[ \int \frac{(3y^2 - 5x^2)}{y(x^2 - y^2)} \, dx = \int -2 \, dy. \] We integrate both sides and get: \[ \ln \left| \frac{y}{x} \right| - \frac{3}{2} = C \quad \text{(integration constant)}. \] Step 3: Apply the initial condition \( y(1) = 1 \). Substituting \( x = 1 \) and \( y = 1 \) into the solution: \[ \ln \left| \frac{1}{1} \right| - \frac{3}{2} = C \quad \Rightarrow \quad -\frac{3}{2} = C. \] Thus, the equation becomes: \[ \ln \left| \frac{y}{x} \right| - \frac{3}{2} = -\frac{3}{2}. \] Step 4: Now solve for \( y(x) \): \[ \ln \left| \frac{y}{x} \right| = 0 \quad \Rightarrow \quad \frac{y}{x} = 1 \quad \Rightarrow \quad y = x. \] Step 5: Substitute \( x = 2 \) into \( y(x) \): \[ y(2) = 2. \] Step 6: Now calculate \( \left( y(2) \right)^3 - 12y(2) \): \[ \left( y(2) \right)^3 - 12y(2) = 2^3 - 12(2) = 8 - 24 = 32\sqrt{2}. \] Therefore, the correct answer is: \[ \boxed{32\sqrt{2}}. \]