Let's solve the given differential equation step by step as per Chegg guidelines: Given: \[ \frac{dy}{dx} = \frac{6x^2}{6y^2 - 3}, \quad y(2) = 0 \] --- ### **Step 1: Separate variables and integrate** **Explanation:** Rearrange the equation to get all \( y \) terms on one side and all \( x \) terms on the other, then integrate both sides. \[ \frac{dy}{dx} = \frac{6x^2}{6y^2 - 3} \] Multiply both sides by \( (6y^2 - 3) \): \[ (6y^2 - 3)\frac{dy}{dx} = 6x^2 \] Now, rewrite as: \[ (6y^2 - 3)dy = 6x^2 dx \] Integrate both sides: \[ \int (6y^2 - 3)dy = \int 6x^2 dx \] \[ 6\int y^2 dy - 3\int dy = 6\int x^2 dx \] \[ 6 \cdot \frac{y^3}{3} - 3y = 6 \cdot \frac{x^3}{3} + C \] \[ 2y^3 - 3y = 2x^3 + C \] --- ### **Step 2: Apply the initial condition to find \( C \)** **Explanation:** Substitute \( x = 2 \) and \( y = 0 \) to solve for \( C \). Given \( y(2) = 0 \): \[ 2(0)^3 - 3(0) = 2(2)^3 + C \] \[ 0 = 2 \times 8 + C \] \[ 0 = 16 + C \implies C = -16 \] So the solution (as an equation) is: \[ \boxed{2y^3 - 3y = 2x^3 - 16} \] --- ### **Domain of Validity** **Explanation:** The original equation is only defined when the denominator is not zero: \( 6y^2 - 3 \neq 0 \implies y^2 \neq \frac{1}{2} \implies y \neq \pm \frac{1}{\sqrt{2}} \). So, the solution is valid for values of \( x \) such that \( y(x) \neq \pm \frac{1}{\sqrt{2}} \). Find the \( x \) values where \( y = \pm \frac{1}{\sqrt{2}} \): Plug \( y = \frac{1}{\sqrt{2}} \) into the solution: \[ 2\left( \frac{1}{\sqrt{2}} \right)^3 - 3\left( \frac{1}{\sqrt{2}} \right) = 2x^3 - 16 \] Calculate: \[ \left( \frac{1}{\sqrt{2}} \right)^3 = \frac{1}{2\sqrt{2}} \] So, \[ 2 \cdot \frac{1}{2\sqrt{2}} - 3 \cdot \frac{1}{\sqrt{2}} = 2x^3 - 16 \] \[ \frac{1}{\sqrt{2}} - \frac{3}{\sqrt{2}} = 2x^3 - 16 \] \[ -\frac{2}{\sqrt{2}} = 2x^3 - 16 \] \[ -\sqrt{2} = 2x^3 - 16 \] \[ 2x^3 = 16 - \sqrt{2} \] \[ x^3 = \frac{16 - \sqrt{2}}{2} \] \[ x_1 = \sqrt[3]{\frac{16 - \sqrt{2}}{2}} \] Similarly for \( y = -\frac{1}{\sqrt{2}} \): \[ 2\left( -\frac{1}{\sqrt{2}} \right)^3 - 3\left( -\frac{1}{\sqrt{2}} \right) = 2x^3 - 16 \] \[ 2 \cdot -\frac{1}{2\sqrt{2}} + \frac{3}{\sqrt{2}} = 2x^3 - 16 \] \[ -\frac{1}{\sqrt{2}} + \frac{3}{\sqrt{2}} = 2x^3 - 16 \] \[ \frac{2}{\sqrt{2}} = 2x^3 - 16 \] \[ \sqrt{2} = 2x^3 - 16 \] \[ 2x^3 = 16 + \sqrt{2} \] \[ x^3 = \frac{16 + \sqrt{2}}{2} \] \[ x_2 = \sqrt[3]{\frac{16 + \sqrt{2}}{2}} \] **Domain:** \[ x_1 < x < x_2 \] \[ \boxed{ \sqrt[3]{\frac{16 - \sqrt{2}}{2}} < x < \sqrt[3]{\frac{16 + \sqrt{2}}{2}} } \] --- ## **Final Answers:** ### **A)** \[ \boxed{2y^3 - 3y = 2x^3 - 16} \] ### **B)** \[ \boxed{ \sqrt[3]{\frac{16 - \sqrt{2}}{2}} < x < \sqrt[3]{\frac{16 + \sqrt{2}}{2}} } \]

Certainly! Here's a detailed explanation in words for the second step: --- **Step 2 Explanation: Applying the initial condition to find the constant \( C \)** In the first step, we derived a general solution involving an arbitrary constant \( C \). To find this specific constant, we use the initial condition provided in the problem, which states that when \( x = 2 \), the value of \( y \) is 0. This means we substitute \( x = 2 \) and \( y = 0 \) into our general solution equation: \[ 2y^3 - 3y = 2x^3 + C \] Plugging in the values: \[ 2 \times 0^3 - 3 \times 0 = 2 \times 2^3 + C \] Simplifies to: \[ 0 = 2 \times 8 + C \] \[ 0 = 16 + C \] From this, we solve for \( C \): \[ C = -16 \] This step ensures the particular solution matches the initial condition, giving us a fully specified equation: \[ 2y^3 - 3y = 2x^3 - 16 \] This is the explicit relation between \( x \) and \( y \) that satisfies both the differential equation and the initial condition.