$(x^2-4)y' - 2xy + 2x(4-x^2)^2 = 0$, $x > 2$,
passes through the point (3, 15), then the local maximum value of $f$ is _____
Solving ODE: Finding Local Maximum of $f(x)$
The problem asks for the local maximum value of the solution curve $y = f(x)$ to the differential equation $(x^2-4)y' - 2xy + 2x(4-x^2)^2 = 0$, given that $x > 2$ and the curve passes through the point (3, 15).
Step 1: Rearrange the Differential Equation
First, rewrite the given differential equation into the standard linear form $y' + P(x)y = Q(x)$.
Original equation: $(x^2-4)y' - 2xy + 2x(4-x^2)^2 = 0$
Isolate the $y'$ term: $(x^2-4)y' - 2xy = -2x(4-x^2)^2$
Divide by $(x^2-4)$ (since $x > 2$, $x^2-4 \neq 0$):
$ y' - \frac{2x}{x^2-4}y = \frac{-2x(4-x^2)^2}{x^2-4} $Simplify the right side using $(4-x^2)^2 = (-(x^2-4))^2 = (x^2-4)^2$:
$ y' - \frac{2x}{x^2-4}y = \frac{-2x(x^2-4)^2}{x^2-4} $ $ y' - \frac{2x}{x^2-4}y = -2x(x^2-4) $This is now in the standard form with $P(x) = -\frac{2x}{x^2-4}$ and $Q(x) = -2x(x^2-4)$.
Step 2: Calculate the Integrating Factor
The integrating factor $\mu(x)$ is given by $e^{\int P(x) dx}$.
$ \int P(x) dx = \int -\frac{2x}{x^2-4} dx $Let $u = x^2-4$, then $du = 2x dx$. The integral becomes:
$ \int -\frac{du}{u} = -\ln|u| = -\ln|x^2-4| $Since $x > 2$, $x^2-4 > 0$, so $|x^2-4| = x^2-4$. The integral is $-\ln(x^2-4)$.
The integrating factor is:
$ \mu(x) = e^{-\ln(x^2-4)} = e^{\ln((x^2-4)^{-1})} = (x^2-4)^{-1} = \frac{1}{x^2-4} $Step 3: Find the General Solution
Multiply the standard form equation by $\mu(x)$:
$ \frac{1}{x^2-4} \left( y' - \frac{2x}{x^2-4}y \right) = \frac{1}{x^2-4} \left( -2x(x^2-4) \right) $ $ \left( \frac{y}{x^2-4} \right)' = -2x $Integrate both sides with respect to $x$:
$ \int \left( \frac{y}{x^2-4} \right)' dx = \int -2x dx $ $ \frac{y}{x^2-4} = -x^2 + C $The general solution is:
$ y = (C - x^2)(x^2-4) $Step 4: Determine the Particular Solution
Use the condition that the solution curve passes through the point (3, 15).
Substitute $x=3$ and $y=15$ into the general solution:
$ 15 = (C - 3^2)(3^2-4) $ $ 15 = (C - 9)(9-4) $ $ 15 = (C - 9)(5) $Solve for C:
$ \frac{15}{5} = C - 9 $ $ 3 = C - 9 $ $ C = 12 $The particular solution is $f(x) = (12 - x^2)(x^2-4)$.
Step 5: Find the Local Maximum Value
Expand $f(x)$:
$ f(x) = 12x^2 - 48 - x^4 + 4x^2 $ $ f(x) = -x^4 + 16x^2 - 48 $To find the local maximum, calculate the first derivative $f'(x)$ and set it to zero.
$ f'(x) = \frac{d}{dx}(-x^4 + 16x^2 - 48) = -4x^3 + 32x $Set $f'(x) = 0$:
$ -4x^3 + 32x = 0 $ $ -4x(x^2 - 8) = 0 $The critical points are $x=0$, $x=\sqrt{8}=2\sqrt{2}$, and $x=-\sqrt{8}=-2\sqrt{2}$.
Consider the domain $x > 2$. The only critical point in the domain is $x = 2\sqrt{2}$ (since $2\sqrt{2} \approx 2.828$).
Use the second derivative test to confirm if it's a maximum.
$ f''(x) = \frac{d}{dx}(-4x^3 + 32x) = -12x^2 + 32 $Evaluate $f''(x)$ at $x = 2\sqrt{2}$:
$ f''(2\sqrt{2}) = -12(2\sqrt{2})^2 + 32 = -12(8) + 32 = -96 + 32 = -64 $Since $f''(2\sqrt{2}) < 0$, the function $f(x)$ has a local maximum at $x = 2\sqrt{2}$.
Calculate the local maximum value by substituting $x = 2\sqrt{2}$ into $f(x)$. Note that $x^2 = (2\sqrt{2})^2 = 8$.
$ f(2\sqrt{2}) = (12 - (2\sqrt{2})^2)((2\sqrt{2})^2 - 4) $ $ f(2\sqrt{2}) = (12 - 8)(8 - 4) $ $ f(2\sqrt{2}) = (4)(4) $ $ f(2\sqrt{2}) = 16 $The local maximum value of $f$ is 16.
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Let $[\cdot]$ be the greatest integer function. If $\alpha = \int_{0}^{64} (x^{1/3} - [x^{1/3}]) dx$, then $\frac{1}{\pi} \int_{0}^{\alpha \pi} \left(\frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta}\right) d\theta$ is equal to ________.
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