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 given differential equation is: \[ \frac{dy}{dx} - \frac{y}{x} + \csc\left(\frac{y}{x}\right) = 0 \] We can rearrange the terms: \[ \frac{dy}{dx} = \frac{y}{x} - \csc\left(\frac{y}{x}\right) \] Now, we use the substitution \( z = \frac{y}{x} \), so that \( y = xz \). Differentiating \( y = xz \) with respect to \( x \), we get: \[ \frac{dy}{dx} = z + x\frac{dz}{dx} \] Substitute this into the original equation: \[ z + x\frac{dz}{dx} = z - \csc(z) \] Simplifying: \[ x\frac{dz}{dx} = -\csc(z) \] Now, we separate the variables: \[ \frac{dz}{\csc(z)} = -\frac{dx}{x} \] Integrating both sides: \[ \int \frac{dz}{\csc(z)} = \int -\frac{dx}{x} \] The integral of \( \frac{1}{\csc(z)} \) is \( \ln|\sin(z)| \), and the integral of \( \frac{1}{x} \) is \( \ln|x| \). Thus, we get: \[ \ln|\sin(z)| = -\ln|x| + C \] Exponentiating both sides: \[ |\sin(z)| = \frac{C}{|x|} \] Since \( z = \frac{y}{x} \), we substitute back to get: \[ |\sin\left(\frac{y}{x}\right)| = \frac{C}{|x|} \] Now, we apply the initial condition \( y = 0 \) when \( x = 1 \): \[ |\sin\left(\frac{0}{1}\right)| = \frac{C}{1} \] This gives \( C = 0 \), so the solution is: \[ \sin\left(\frac{y}{x}\right) = 0 \] Thus, the particular solution is: \[ \frac{y}{x} = n\pi \quad \text{(for some integer \( n \))} \] Since \( y = 0 \) when \( x = 1 \), we have \( n = 0 \), and therefore the solution is: \[ y = 0 \]