Question

Find the particular solution of the differential equation \( \frac{dy}{dx} - \frac{y}{x} + \csc\left(\frac{y}{x}\right) = 0 \); given that \( y = 0 \), when \( x = 1 \).

Hint:

When solving a differential equation using substitution, always check the initial conditions to find the specific solution. The key steps involve separating variables and performing integration correctly.

Answer 1

The differential equation is given as: \[ \frac{dy}{dx} - \frac{y}{x} + \csc\left(\frac{y}{x}\right) = 0 \] Rearranging yields: \[ \frac{dy}{dx} = \frac{y}{x} - \csc\left(\frac{y}{x}\right) \] Let \( z = \frac{y}{x} \), implying \( y = xz \). Differentiating \( y \) with respect to \( x \) gives \( \frac{dy}{dx} = z + x\frac{dz}{dx} \). Substituting this into the rearranged equation: \[ z + x\frac{dz}{dx} = z - \csc(z) \] This simplifies to: \[ x\frac{dz}{dx} = -\csc(z) \] Separating variables: \[ \frac{dz}{\csc(z)} = -\frac{dx}{x} \] Integrating both sides: \[ \int \sin(z) \, dz = \int -\frac{1}{x} \, dx \] The integration yields: \[ \ln|\sin(z)| = -\ln|x| + C \] Exponentiating both sides: \[ |\sin(z)| = \frac{C}{|x|} \] Substituting back \( z = \frac{y}{x} \): \[ |\sin\left(\frac{y}{x}\right)| = \frac{C}{|x|} \] Applying the initial condition \( y = 0 \) when \( x = 1 \): \[ |\sin\left(\frac{0}{1}\right)| = \frac{C}{1} \] This results in \( C = 0 \), so the equation becomes: \[ \sin\left(\frac{y}{x}\right) = 0 \] This implies: \[ \frac{y}{x} = n\pi \quad \text{(where \( n \) is an integer)} \] Using the initial condition again, \( \frac{0}{1} = n\pi \), which means \( n = 0 \). Therefore, the particular solution is: \[ y = 0 \]