Key Concepts and Formulas

• Area Under a Curve: The area under the curve $y=f(x)$ from $x=a$ to $x=b$ is given by $\int_a^b f(x) dx$.
• Leibniz Rule: $\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$.
• Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ has the solution $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C$, where $\text{I.F.} = e^{\int P(x) dx}$.

Step-by-Step Solution

Step 1: Setting up the Initial Equation

The area under the curve $y=y(x)$ from $x=3$ to $x=x$ is given as $(\frac{y}{x})^3$. We express this as an integral equation:

$\int_3^x y(t) dt = \left(\frac{y(x)}{x}\right)^3$

To simplify differentiation, we multiply both sides by $x^3$:

$x^3 \int_3^x y(t) dt = y^3(x) \quad (*)$

Step 2: Differentiating using Leibniz Rule

We differentiate both sides of equation $(*)$ with respect to $x$ using Leibniz rule and the product rule.

$\frac{d}{dx} \left(x^3 \int_3^x y(t) dt\right) = \frac{d}{dx} (y^3)$

Applying the product rule and Leibniz rule, we get:

$3x^2 \int_3^x y(t) dt + x^3 y(x) = 3y^2 \frac{dy}{dx}$

Now, substitute $\int_3^x y(t) dt = \frac{y^3}{x^3}$ from equation $(*)$:

$3x^2 \left(\frac{y^3}{x^3}\right) + x^3 y = 3y^2 \frac{dy}{dx}$

Simplifying, we get:

$\frac{3y^3}{x} + x^3 y = 3y^2 \frac{dy}{dx}$

Divide by $y$ (since $y \neq 0$):

$\frac{3y^2}{x} + x^3 = 3y \frac{dy}{dx}$

Rearrange to isolate the derivative term:

$3y \frac{dy}{dx} = \frac{3y^2}{x} + x^3$

$3xy \frac{dy}{dx} = 3y^2 + x^4$

Step 3: Solving the Differential Equation using Substitution

Let $t = y^2$. Then $\frac{dt}{dx} = 2y \frac{dy}{dx}$, so $y \frac{dy}{dx} = \frac{1}{2} \frac{dt}{dx}$. Substituting these into the differential equation, we get:

$3x \left(\frac{1}{2} \frac{dt}{dx}\right) = 3t + x^4$

$\frac{3}{2} x \frac{dt}{dx} = 3t + x^4$

Divide by $\frac{3}{2}x$:

$\frac{dt}{dx} = \frac{2}{3x} (3t + x^4)$

$\frac{dt}{dx} = \frac{2t}{x} + \frac{2x^3}{3}$

Rearrange into the standard linear differential equation form:

$\frac{dt}{dx} - \frac{2}{x} t = \frac{2x^3}{3}$

Here, $P(x) = -\frac{2}{x}$ and $Q(x) = \frac{2x^3}{3}$. The integrating factor is:

$\text{I.F.} = e^{\int P(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2 \ln|x|} = e^{\ln(x^{-2})} = x^{-2} = \frac{1}{x^2}$

The general solution is:

$t \cdot \left(\frac{1}{x^2}\right) = \int \left(\frac{2x^3}{3}\right) \cdot \left(\frac{1}{x^2}\right) dx + C$

$\frac{t}{x^2} = \int \frac{2x}{3} dx + C$

$\frac{t}{x^2} = \frac{x^2}{3} + C$

Substituting back $t=y^2$, we have:

$\frac{y^2}{x^2} = \frac{x^2}{3} + C$

$y^2 = \frac{x^4}{3} + Cx^2$

Step 4: Applying the Initial Condition

The curve passes through $(3,3)$. Substituting $x=3$ and $y=3$ into the general solution:

$3^2 = \frac{3^4}{3} + C(3^2)$

$9 = \frac{81}{3} + 9C$

$9 = 27 + 9C$

$9C = -18$

$C = -2$

The particular solution is:

$y^2 = \frac{x^4}{3} - 2x^2$

Step 5: Finding $\alpha$

The curve also passes through $(\alpha, 6\sqrt{10})$. Substituting $x=\alpha$ and $y=6\sqrt{10}$:

$(6\sqrt{10})^2 = \frac{\alpha^4}{3} - 2\alpha^2$

$360 = \frac{\alpha^4}{3} - 2\alpha^2$

Multiply by 3:

$1080 = \alpha^4 - 6\alpha^2$

$\alpha^4 - 6\alpha^2 - 1080 = 0$

Let $A = \alpha^2$:

$A^2 - 6A - 1080 = 0$

Using the quadratic formula:

$A = \frac{6 \pm \sqrt{36 - 4(-1080)}}{2} = \frac{6 \pm \sqrt{36 + 4320}}{2} = \frac{6 \pm \sqrt{4356}}{2} = \frac{6 \pm 66}{2}$

So $A = \frac{6+66}{2} = \frac{72}{2} = 36$ or $A = \frac{6-66}{2} = \frac{-60}{2} = -30$. Since $A = \alpha^2$, $A$ must be non-negative. Thus, $A=36$.

$\alpha^2 = 36$

$\alpha = \pm 6$

Since the point is in the first quadrant, $\alpha > 0$, so $\alpha = 6$.

Common Mistakes & Tips

• Remember to use the product rule when differentiating a product of functions.
• Carefully apply the Leibniz rule, paying attention to the limits of integration.
• When solving linear differential equations, double-check the calculation of the integrating factor.

Summary

We started by setting up an integral equation based on the area under the curve. We then differentiated this equation using Leibniz rule and the product rule to obtain a differential equation. By using a substitution, we transformed the differential equation into a linear form and solved it using an integrating factor. Finally, we applied the given points to find the particular solution and the value of $\alpha$.

The final answer is \boxed{6}.