Given the differential equation:
\[ (x - y) \, dy - (x + y) \, dx = 0 \]
Rearrange the equation:
\[ (x - y) \, dy = (x + y) \, dx \]
Now, divide both sides by \( (x - y)(x + y) \):
\[ \frac{dy}{dx} = \frac{x + y}{x - y} \]
This is a **homogeneous equation** and can be solved using the substitution \( v = \frac{y}{x} \), where \( y = vx \). Hence, \( dy = v dx + x dv \).
Substitute into the equation:
\[ \frac{v dx + x dv}{dx} = \frac{x + vx}{x - vx} \] \[ v + x \frac{dv}{dx} = \frac{x(1 + v)}{x(1 - v)} \] \[ v + x \frac{dv}{dx} = \frac{1 + v}{1 - v} \]
Now, isolate the terms with \( dv \) on one side:
\[ x \frac{dv}{dx} = \frac{1 + v}{1 - v} - v \] Simplifying the right-hand side: \[ x \frac{dv}{dx} = \frac{1 + v - v(1 - v)}{1 - v} \] \[ x \frac{dv}{dx} = \frac{1 + v - v + v^2}{1 - v} \] \[ x \frac{dv}{dx} = \frac{1 + v^2}{1 - v} \] Now, the equation is ready to be solved by integrating both sides.
Continue solving by simplifying and performing the necessary integrations to obtain the solution of the differential equation.
Let \( y = y(x) \) be the solution of the differential equation \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] such that \( y(0) = \frac{5}{4} \). Then \[ 12 \left( y\left( \frac{\pi}{4} \right) - e^{-2} \right) \] is equal to _____.