Key Concepts and Formulas

• Homogeneous Differential Equation: A differential equation of the form $\frac{dy}{dx} = f(x,y)$ is homogeneous if $f(\lambda x, \lambda y) = f(x,y)$ for any non-zero $\lambda$.
• Substitution Method: To solve a homogeneous differential equation, substitute $y = vx$, which implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
• Separation of Variables: After the substitution, the equation should be separable, meaning it can be written in the form $g(v)dv = h(x)dx$. Then integrate both sides.

Step-by-Step Solution

Step 1: Identify and Rewrite the Differential Equation

The given differential equation is: $(x^2+y^2)dx - 5xydy = 0$ We rewrite it in the form $\frac{dy}{dx} = f(x,y)$: $(x^2+y^2)dx = 5xydy$ $\frac{dy}{dx} = \frac{x^2+y^2}{5xy}$ This confirms that the equation is homogeneous, as $f(x,y) = \frac{x^2+y^2}{5xy}$ and $f(\lambda x, \lambda y) = \frac{(\lambda x)^2+(\lambda y)^2}{5(\lambda x)(\lambda y)} = \frac{\lambda^2(x^2+y^2)}{5\lambda^2 xy} = \frac{x^2+y^2}{5xy} = f(x,y)$.

Step 2: Apply the Substitution

To solve the homogeneous differential equation, we use the substitution $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting into the equation, we get: $v + x\frac{dv}{dx} = \frac{x^2 + (vx)^2}{5x(vx)}$ $v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{5vx^2}$ $v + x\frac{dv}{dx} = \frac{x^2(1 + v^2)}{5vx^2}$ $v + x\frac{dv}{dx} = \frac{1 + v^2}{5v}$

Step 3: Separate the Variables

Now, we isolate the terms involving $v$ and $x$: $x\frac{dv}{dx} = \frac{1 + v^2}{5v} - v$ $x\frac{dv}{dx} = \frac{1 + v^2 - 5v^2}{5v}$ $x\frac{dv}{dx} = \frac{1 - 4v^2}{5v}$ Separate the variables: $\frac{5v}{1 - 4v^2} dv = \frac{dx}{x}$

Step 4: Integrate Both Sides

Integrate both sides of the equation: $\int \frac{5v}{1 - 4v^2} dv = \int \frac{dx}{x}$ For the integral on the left side, let $u = 1 - 4v^2$. Then $du = -8v dv$, so $v dv = -\frac{1}{8}du$. Therefore, $\int \frac{5v}{1 - 4v^2} dv = \int \frac{5}{u} \left(-\frac{1}{8}\right) du = -\frac{5}{8} \int \frac{1}{u} du = -\frac{5}{8} \ln|u| + C_1 = -\frac{5}{8} \ln|1 - 4v^2| + C_1$ The integral on the right side is: $\int \frac{dx}{x} = \ln|x| + C_2$ So we have: $-\frac{5}{8} \ln|1 - 4v^2| = \ln|x| + C$ where $C = C_2 - C_1$.

Step 5: Substitute Back and Simplify

Substitute $v = \frac{y}{x}$ back into the equation: $-\frac{5}{8} \ln\left|1 - 4\left(\frac{y}{x}\right)^2\right| = \ln|x| + C$ $-\frac{5}{8} \ln\left|1 - \frac{4y^2}{x^2}\right| = \ln|x| + C$ $-\frac{5}{8} \ln\left|\frac{x^2 - 4y^2}{x^2}\right| = \ln|x| + C$ Multiply by $-\frac{8}{5}$: $\ln\left|\frac{x^2 - 4y^2}{x^2}\right| = -\frac{8}{5} \ln|x| + C'$ where $C' = -\frac{8}{5}C$. Exponentiate both sides: $\left|\frac{x^2 - 4y^2}{x^2}\right| = e^{-\frac{8}{5} \ln|x| + C'} = e^{C'} e^{\ln|x|^{-\frac{8}{5}}} = e^{C'} |x|^{-\frac{8}{5}}$ $\left|\frac{x^2 - 4y^2}{x^2}\right| = A |x|^{-\frac{8}{5}}$ where $A = e^{C'} > 0$. $|x^2 - 4y^2| = A |x|^{-\frac{8}{5}} |x^2| = A |x|^{\frac{2}{5}}$ $|x^2 - 4y^2| = A x^{\frac{2}{5}}$

Step 6: Apply the Initial Condition

Given $y(1) = 0$, substitute $x = 1$ and $y = 0$: $|1^2 - 4(0)^2| = A (1)^{\frac{2}{5}}$ $|1| = A$ $A = 1$ Thus, the solution is: $|x^2 - 4y^2| = x^{\frac{2}{5}}$ Raise both sides to the power of 5: $|x^2 - 4y^2|^5 = (x^{\frac{2}{5}})^5$ $|x^2 - 4y^2|^5 = x^2$

Common Mistakes & Tips

• Sign Errors: Be careful when integrating and substituting back. Sign errors are common and can lead to incorrect results.
• Constant of Integration: Don't forget the constant of integration after each integration. It is crucial for finding the particular solution that satisfies the initial condition.
• Absolute Values: Remember to use absolute values inside logarithms to ensure the function is defined for all possible values.

Summary

We solved the homogeneous differential equation $(x^2+y^2)dx - 5xydy = 0$ with the initial condition $y(1) = 0$. We used the substitution $y = vx$, separated the variables, integrated both sides, applied the initial condition to find the constant of integration, and simplified the expression. The final solution is $|x^2 - 4y^2|^5 = x^2$.

Final Answer

The final answer is \boxed{\left|x^2-4 y^2\right|^5=x^2}, which corresponds to option (A).