Question

The solution of the differential equation \( (x^2 + y^2) dx - 5xy \, dy = 0, \, y(1) = 0 \), is:

• A. \(|x^2 - 4y^2|^5 = x^2\)
• B. \(|x^2 - 2y^2|^6 = x\)
• C. \(|x^2 - 4y^2|^6 = x\)
• D. \(|x^2 - 2y^2|^5 = x^2\)

Answer 1

The Correct Option is A

The given differential equation is:

\((x^2 + y^2)dx - 5xy\,dy = 0.\)

Step 1: Rewrite in terms of \(\frac{dy}{dx}\)

\(\frac{dy}{dx} = \frac{x^2 + y^2}{5xy}.\)

Step 2: Substitution

Let \(y = vx\), so \(\frac{dy}{dx} = v + x\frac{dv}{dx}\). Substitute into the equation:

\(v + x\frac{dv}{dx} = \frac{x^2 + (vx)^2}{5x(vx)}.\)

Simplify:

\(v + x\frac{dv}{dx} = \frac{1 + v^2}{5v}.\)

Simplify further:

\(x\frac{dv}{dx} = \frac{1 + v^2}{5v} - v.\)

\(x\frac{dv}{dx} = \frac{1 + v^2 - 5v^2}{5v}.\)

\(x\frac{dv}{dx} = \frac{1 - 4v^2}{5v}.\)

Step 3: Separate variables

\(v\,dv = \frac{dx}{5x(1 - 4v^2)}.\)

Step 4: Solve the integral

Let \(1 - 4v^2 = t\), so \(-8v\,dv = dt\). The left-hand side becomes:

\(\int \frac{v\,dv}{1 - 4v^2} = \int \frac{dx}{5x}.\)

Integrate both sides:

\(-\frac{1}{8} \ln|t| = \frac{1}{5} \ln|x| + \ln C.\)

Substitute \(t = 1 - 4v^2\):

\(-\frac{1}{8} \ln|1 - 4v^2| = \frac{1}{5} \ln|x| + \ln C.\)

Simplify:

\(\ln|x^8| + \ln|1 - 4v^2|^5 = \ln C.\)

\(x^8 |1 - 4v^2|^5 = C.\)

Step 5: Substitute back \(v = \frac{y}{x}\)

\(x^8 |1 - 4\left(\frac{y}{x}\right)^2|^5 = C.\)

\(|x^2 - 4y^2|^5 = Cx^2.\)

Step 6: Apply the initial condition

Given \(y(1) = 0\):

\(|1^2 - 4(0)^2|^5 = C(1^2).\)

\(C = 1.\)

Thus, the solution is:

\(|x^2 - 4y^2|^5 = x^2.\)

Final Answer: Option (1).