Key Concepts and Formulas

• Homogeneous Differential Equation: A differential equation of the form $\frac{dy}{dx} = f(x, y)$ is homogeneous if $f(tx, ty) = f(x, y)$ for all $t$. This means it can be written as $\frac{dy}{dx} = g\left(\frac{y}{x}\right)$.
• Substitution for Homogeneous Equations: To solve a homogeneous differential equation, substitute $y = vx$, which implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
• Integral Formula: $\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln|x + \sqrt{x^2 + a^2}| + C$

Step-by-Step Solution

Step 1: Identify and Transform the Differential Equation into Standard Homogeneous Form

We are given the differential equation: $x\frac{dy}{dx} - y = \sqrt{y^2 + 16x^2}$ Our goal is to isolate $\frac{dy}{dx}$ and express the right-hand side as a function of $\frac{y}{x}$.

• Isolate $\frac{dy}{dx}$: Add $y$ to both sides: $x\frac{dy}{dx} = y + \sqrt{y^2 + 16x^2}$
• Divide by $x$: Divide both sides by $x$, assuming $x \neq 0$: $\frac{dy}{dx} = \frac{y}{x} + \frac{\sqrt{y^2 + 16x^2}}{x}$
• Express the radical term as a function of $\frac{y}{x}$: Since the initial condition $y(1)=3$ implies $x$ is near 1, we can assume $x>0$. Therefore, $x = \sqrt{x^2}$. $\frac{dy}{dx} = \frac{y}{x} + \sqrt{\frac{y^2 + 16x^2}{x^2}}$
• Simplify the terms inside the square root: $\frac{dy}{dx} = \frac{y}{x} + \sqrt{\frac{y^2}{x^2} + 16}$ $\frac{dy}{dx} = \frac{y}{x} + \sqrt{\left(\frac{y}{x}\right)^2 + 16}$ This is now in the form $\frac{dy}{dx} = g\left(\frac{y}{x}\right)$, confirming it's a homogeneous differential equation.

Step 2: Apply the Substitution for Homogeneous Equations

Let $y = vx$. Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.

• Substitute $y=vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ into the transformed differential equation: $v + x\frac{dv}{dx} = v + \sqrt{v^2 + 16}$
• Simplify the equation: $x\frac{dv}{dx} = \sqrt{v^2 + 16}$

Step 3: Separate Variables

Now we separate the variables to get all $v$ terms with $dv$ and all $x$ terms with $dx$.

• Rearrange terms: Divide both sides by $\sqrt{v^2 + 16}$ and $x$: $\frac{dv}{\sqrt{v^2 + 16}} = \frac{dx}{x}$

Step 4: Integrate Both Sides

Integrate both sides of the separated equation.

• Left side integral: $\int \frac{dv}{\sqrt{v^2 + 16}} = \ln\left|v + \sqrt{v^2 + 16}\right| + C_1$
• Right side integral: $\int \frac{dx}{x} = \ln|x| + C_2$
• Combine and add constant of integration: $\ln\left|v + \sqrt{v^2 + 16}\right| = \ln|x| + \ln|C|$

Step 5: Simplify and Substitute Back $v = \frac{y}{x}$

• Use logarithm properties: $\ln\left|v + \sqrt{v^2 + 16}\right| = \ln|Cx|$
• Exponentiate both sides: $v + \sqrt{v^2 + 16} = Cx$
• Substitute back $v = \frac{y}{x}$: $\frac{y}{x} + \sqrt{\left(\frac{y}{x}\right)^2 + 16} = Cx$
• Simplify the term under the square root: $\frac{y}{x} + \sqrt{\frac{y^2 + 16x^2}{x^2}} = Cx$
• Simplify the square root: Since $x>0$, $\sqrt{x^2}=x$. $\frac{y}{x} + \frac{\sqrt{y^2 + 16x^2}}{x} = Cx$
• Multiply by x: $y + \sqrt{y^2 + 16x^2} = Cx^2$

Step 6: Use the Initial Condition to Find the Constant $C$

We are given $y(1) = 3$. Substitute $x=1$ and $y=3$ into the equation: $3 + \sqrt{3^2 + 16(1)^2} = C(1)^2$ $3 + \sqrt{9 + 16} = C$ $3 + \sqrt{25} = C$ $3 + 5 = C$ $C = 8$

Step 7: Write the Particular Solution

Substitute $C=8$ back into the equation: $y + \sqrt{y^2 + 16x^2} = 8x^2$

Step 8: Calculate $y(2)$

Substitute $x=2$ into the particular solution: $y + \sqrt{y^2 + 16(2)^2} = 8(2)^2$ $y + \sqrt{y^2 + 64} = 32$

• Isolate the square root: $\sqrt{y^2 + 64} = 32 - y$
• Square both sides: $y^2 + 64 = (32 - y)^2$ $y^2 + 64 = 1024 - 64y + y^2$
• Solve for $y$: $64 = 1024 - 64y$ $64y = 960$ $y = \frac{960}{64} = 15$
• Check for extraneous solutions: Since $32-y = 32-15 = 17 > 0$, the solution is valid.

Thus, $y(2) = 15$.

Common Mistakes & Tips

• Sign of x: When taking $x$ inside the square root, remember to consider the sign of $x$. In this case, the initial condition implied $x>0$.
• Extraneous Solutions: Always check for extraneous solutions when squaring both sides of an equation.
• Logarithm Properties: Remember and apply logarithm properties correctly to simplify the equations.

Summary

We solved the given homogeneous differential equation by using the substitution $y=vx$. This transformed the equation into a separable form, which we integrated and solved for $y$ using the initial condition. Finally, we calculated $y(2)$, which is 15.

The final answer is \boxed{15}, which corresponds to option (A).