Question

The solution of \( e^{y-x} \frac{dy}{dx} = \frac{y(\sin x + \cos x)}{1 + y \log y} \)

Answer 1

Solution to the Differential Equation
Given the differential equation: \[ e^{y-x} \frac{dy}{dx} = \frac{y(\sin x + \cos x)}{1 + y \log y} \] Rearrange to separate variables: \[ e^{y-x} \frac{dy}{dx} = \frac{y (\sin x + \cos x)}{1 + y \log y} \]

Simplify and Integrate Both Sides:
We separate the variables to set up the integration: \[ \int (1 + \log y) e^y \, dy = \int (\sin x + \cos x) e^x \, dx \]

Using Integration by Parts:
The left-hand side can be solved using integration by parts. Let's integrate each part:
\[ \int (1 + \log y) e^y \, dy = \log y \cdot e^y - \int \frac{e^y}{y} \, dy \]

Right-hand side Integration:
On the right-hand side: \[ \int (\sin x + \cos x) e^x \, dx = e^x (\sin x - \cos x) \]

Final Solution:
Combining the results from both sides gives us the general solution: \[ \log y \cdot e^y - \int \frac{e^y}{y} \, dy = e^x (\sin x - \cos x) + C \] Where \( C \) is the constant of integration.