Solve the IVP√yy′(x) + y 32x − x = 0 , y(4) = 0 ,using both the Leibniz and the Bernoulli methods.Show that your answers are equivalent.

To solve the initial value problem (IVP) given by the differential equation

$\sqrt{y} y'(x) + y (32x - x) = 0$

with the initial condition $y(4) = 0$, we will utilize both the Leibniz method and the Bernoulli method.

Step 1: Break Down the Problem

1. Rearrange the equation to isolate $y'$.
2. Identify if it fits the Bernoulli differential equation form.
3. Solve using both methods and compare the results.

Step 2: Relevant Concepts

1. Leibniz Method: Often involves integrating directly or using substitutions.
2. Bernoulli's Equation: Has the form $y' + P(x)y = Q(x)y^n$. We should try to express our equation in this form.

Step 3: Analysis and Detail

1. Rearranging the Leibniz Method:

   Starting with the original equation: $\sqrt{y} y' + y (32x - x) = 0$ We can write it as: $y' = -\frac{y (32x - x)}{\sqrt{y}}$ Simplifying gives: $y' = -\frac{31xy}{\sqrt{y}} = -31x \sqrt{y}$

2. Using Bernoulli's Method:

   We need to rewrite it to resemble the Bernoulli form:

   Let $y = v^2$ (where $y\geq0$), then $y' = 2v v'$:

   Substituting this in gives: $2v v' = -31x v \implies v' = -\frac{31x}{2}v$ This is now a linear first-order differential equation.

   Rearranging: $v' + \frac{31x}{2} v = 0$

   This can be solved using integrating factors.

   The integrating factor $\mu(x)$ is: $\mu(x) = e^{\int \frac{31x}{2} dx} = e^{\frac{31}{4}x^2}$

   Multiplying through by $\mu(x)$: $e^{\frac{31}{4}x^2} v' + \frac{31x}{2} e^{\frac{31}{4}x^2} v = 0$

   The left side is the derivative: $\frac{d}{dx}(e^{\frac{31}{4}x^2} v) = 0$

   Integrating gives: $e^{\frac{31}{4}x^2} v = C \implies v = Ce^{-\frac{31}{4}x^2}$

   Re-substituting $y = v^2$: $y = C^2 e^{-\frac{31}{2}x^2}$

3. Applying the Initial Condition:

   From the initial condition $y(4) = 0$: $C^2 e^{-\frac{31}{2} \cdot 4^2} = 0 \implies C = 0$

Step 4: Verify and Summarize

In both approaches, we find that $y \equiv 0$ is the only solution consistent with the initial value.

Final Answer

The solution to the initial value problem is: $\boxed{y(x) = 0}$