Key Concepts and Formulas

• Homogeneous Differential Equations: A differential equation of the form $\frac{dy}{dx} = f(x, y)$ is homogeneous if $f(tx, ty) = f(x, y)$ for all $t$. These are solved using the substitution $y = vx$.
• Separation of Variables: A technique for solving differential equations where terms involving $y$ are separated to one side and terms involving $x$ to the other.
• Integration: Finding the integral of a function.

Step-by-Step Solution

Step 1: Identify the differential equation and check for homogeneity

We are given the differential equation: $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$ Let's check if it's homogeneous by substituting $x$ with $tx$ and $y$ with $ty$: $f(tx, ty) = \frac{(tx)^2 + (ty)^2}{2(tx)(ty)} = \frac{t^2x^2 + t^2y^2}{2t^2xy} = \frac{t^2(x^2 + y^2)}{t^2(2xy)} = \frac{x^2 + y^2}{2xy} = f(x, y)$ Since $f(tx, ty) = f(x, y)$, the differential equation is homogeneous.

Step 2: Apply the substitution $y = vx$

Let $y = vx$. Then, differentiating with respect to $x$, we get: $\frac{dy}{dx} = v + x \frac{dv}{dx}$ Substitute $y = vx$ into the original differential equation: $\frac{dy}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{x^2 + v^2x^2}{2vx^2} = \frac{x^2(1 + v^2)}{2vx^2} = \frac{1 + v^2}{2v}$

Step 3: Substitute $\frac{dy}{dx}$ and solve the resulting differential equation

Now, substitute $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the equation from Step 2: $v + x \frac{dv}{dx} = \frac{1 + v^2}{2v}$ $x \frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}$ Now we have a separable differential equation: $\frac{2v}{1 - v^2} dv = \frac{dx}{x}$

Step 4: Integrate both sides

Integrate both sides of the equation: $\int \frac{2v}{1 - v^2} dv = \int \frac{dx}{x}$ Let $u = 1 - v^2$, then $du = -2v dv$. So, the left side becomes: $-\int \frac{du}{u} = -\ln|u| = -\ln|1 - v^2|$ The right side is: $\int \frac{dx}{x} = \ln|x| + C$ Therefore, we have: $-\ln|1 - v^2| = \ln|x| + C$ $\ln|1 - v^2|^{-1} = \ln|x| + C$ $\ln\left|\frac{1}{1 - v^2}\right| = \ln|x| + C$ Exponentiate both sides: $\frac{1}{|1 - v^2|} = e^{\ln|x| + C} = e^{\ln|x|} \cdot e^C = |x|e^C$ Let $A = \pm e^C$, where the sign absorbs the absolute value on the left side. Then: $\frac{1}{1 - v^2} = Ax$ $1 = Ax(1 - v^2)$ $1 = Ax\left(1 - \frac{y^2}{x^2}\right) = Ax\left(\frac{x^2 - y^2}{x^2}\right)$ $1 = A\left(\frac{x^2 - y^2}{x}\right)$ $x = A(x^2 - y^2)$ $x = Ax^2 - Ay^2$

Step 5: Use the initial condition $y(2) = 0$ to find the constant A

We are given that $y(2) = 0$. Substitute $x = 2$ and $y = 0$ into the equation: $2 = A(2^2 - 0^2) = 4A$ $A = \frac{2}{4} = \frac{1}{2}$

Step 6: Substitute the value of A back into the equation

Substitute $A = \frac{1}{2}$ into the equation $x = A(x^2 - y^2)$: $x = \frac{1}{2}(x^2 - y^2)$ $2x = x^2 - y^2$ $y^2 = x^2 - 2x$

Step 7: Find $y(8)$

We want to find $y(8)$. Substitute $x = 8$ into the equation: $y^2 = (8)^2 - 2(8) = 64 - 16 = 48$ $y = \pm \sqrt{48} = \pm \sqrt{16 \cdot 3} = \pm 4\sqrt{3}$ Since none of the options match $+4\sqrt{3}$, and the answer is $- 4\sqrt 2$, there must be an error. Going back, we made an error in the integration.

Let us go back to the equation $-\ln|1 - v^2| = \ln|x| + C$ Instead of exponentiating, we can write it as: $\ln|1-v^2| + \ln|x| = -C$ $\ln|(1-v^2)x| = -C$ $(1-v^2)x = e^{-C} = K$ $x(1 - \frac{y^2}{x^2}) = K$ $x - \frac{y^2}{x} = K$ $x^2 - y^2 = Kx$

Now, we use the initial condition $y(2) = 0$, so $x = 2$ and $y = 0$: $2^2 - 0^2 = K(2)$ $4 = 2K$ $K = 2$

Substitute $K = 2$ back into the equation: $x^2 - y^2 = 2x$ $y^2 = x^2 - 2x$ Now, substitute $x = 8$: $y^2 = 8^2 - 2(8) = 64 - 16 = 48$ $y = \pm \sqrt{48} = \pm 4\sqrt{3}$ Since we want $y(8)$, and the final answer is $-4\sqrt{2}$, there must be an error.

Let's re-examine the given solution. We have $y^2 = x^2 - 2x$. Let's try to find a value of $x$ where $y = -4\sqrt{2}$. $(-4\sqrt{2})^2 = x^2 - 2x$ $16(2) = x^2 - 2x$ $32 = x^2 - 2x$ $x^2 - 2x - 32 = 0$ $x = \frac{2 \pm \sqrt{4 - 4(-32)}}{2} = \frac{2 \pm \sqrt{4+128}}{2} = \frac{2 \pm \sqrt{132}}{2} = 1 \pm \sqrt{33}$ The question asks for the value of y(8), so let's rewrite the equation as: $y = \pm \sqrt{x^2 - 2x}$ $y' = \pm \frac{2x - 2}{2\sqrt{x^2 - 2x}} = \pm \frac{x-1}{\sqrt{x^2 - 2x}}$ When x = 8, $y' = \pm \frac{7}{\sqrt{48}} = \pm \frac{7}{4\sqrt{3}}$

It appears there is an error in the problem statement or the provided correct answer. Given the initial condition $y(2)=0$ and the differential equation, the derived equation is $y^2 = x^2 - 2x$. Therefore, $y(8) = \pm \sqrt{8^2 - 2(8)} = \pm \sqrt{48} = \pm 4\sqrt{3}$. Since the expected answer is $y(8) = -4\sqrt{2}$, there might be a mistake in the question or the answer. However, based on the given information, the closest answer would be one of the $\pm 4\sqrt{3}$ values. However, the correct answer is given as $-4\sqrt{2}$. Since our solution is logically sound, we will assume there is a problem with the given answer.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs during integration and substitution. A small sign error can lead to a completely wrong answer.
• Absolute Values: Remember to include absolute values when integrating functions like $\frac{1}{x}$, and consider how to remove them appropriately based on the problem context.
• Checking Homogeneity: Always verify that the differential equation is indeed homogeneous before applying the substitution $y = vx$.

Summary

We solved the given homogeneous differential equation using the substitution $y = vx$ and separation of variables. We found the general solution and then used the initial condition $y(2) = 0$ to determine the constant of integration. This gave us the particular solution $y^2 = x^2 - 2x$. Evaluating this at $x = 8$, we find $y(8) = \pm 4\sqrt{3}$. Since the given correct answer is $-4\sqrt{2}$, there is likely an error in the question or the provided answer. Based on our calculations, the closest answer would be $-4\sqrt{3}$.

Final Answer The final answer is \boxed{-4\sqrt{3}}. which is closest to option (A), but not exactly it. There may be an error in the problem or the given answer.