Question

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

Hint:

For homogeneous differential equations, use substitution \( y = v x \) to simplify the equation and separate variables.

Answer 1

Step 1: Rewrite the given differential equation
Rearrange the equation to separate variables: \[ \frac{dy}{dx} = x (e)^y/^x + \frac{y}{x} \] This is a homogeneous differential equation.
Step 2: Substitution
Let: \[ y = v x \quad \Rightarrow \quad \frac{dy}{dx} = v + x \frac{dv}{dx} \] Substitute \( y = v x \) into the equation: \[ v + x \frac{dv}{dx} = x (e)^y/^x + v \] Simplify: \[ x \frac{dv}{dx} = x (e)^y/^x \] Step 3: Solve for \( v \)
Rearrange the equation for integration: \[ \int e^{-v} \, dv = \int \frac{1}{x} \, dx \] Evaluate the integrals: \[ -e^{-v} = \log |x| + c \] Substitute \( v = \frac{y}{x} \): \[ -e^{-\frac{y}{x}} = \log |x| + c \quad \dots {(1)} \] Step 4: Apply the initial condition
We are given that \( y = 1 \) when \( x = 1 \). Substitute \( x = 1 \) and \( y = 1 \) into equation (1): \[ -e^{-\frac{1}{1}} = \log |1| + c \] Since \( \log |1| = 0 \): \[ -e^{-1} = c \] Thus: \[ c = -e^{-1} \] Step 5: Substitute \( c \) back into the solution
Substitute \( c = -e^{-1} \) into equation (1): \[ -e^{-\frac{y}{x}} = \log |x| - e^{-1} \] Rearranging: \[ \log |x| + e^{-\frac{y}{x}} = e^{-1} \] % Final Answer Final Answer: \[ \boxed{\log |x| + e^{-\frac{y}{x}} = e^{-1}} \]