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

Answer: (A)

$|x^2 - 4y^2|^5 = x^2$

Answer 2

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).