Question

The solution of the differential equation \( x \cos y \, dy = (x e^x \log x + e^x) \, dx \) is:

Hint:

When solving differential equations, it’s important to rearrange and integrate both sides carefully. Look for substitutions or simplifications to make the integration easier. For separable differential equations, integrating each side step by step is crucial.

Answer 1

The differential equation is given by: \[x \cos y \, dy = (x e^x \log x + e^x) \, dx.\]This is a separable equation. Rearranging it yields:\[\cos y \, dy = \left( e^x \log x + \frac{e^x}{x} \right) \, dx.\]Step 1: Integrate both sides. Integrating both sides gives:- Left-hand side: \[\int \cos y \, dy = \sin y + C_1.\]- Right-hand side: \[\int \left( e^x \log x + \frac{e^x}{x} \right) dx.\]Step 2: Solve the right-hand side integral. The integral on the right-hand side can be split: \[\int e^x \log x \, dx + \int \frac{e^x}{x} \, dx.\]- For \( \int e^x \log x \, dx \), integration by parts is used. Let \( u = \log x \) (\( du = \frac{1}{x} dx \)) and \( dv = e^x dx \) (\( v = e^x \)). Applying the formula \( \int u dv = uv - \int v du \): \[ \int e^x \log x \, dx = e^x \log x - \int e^x \frac{1}{x} dx. \]- The integral \( \int \frac{e^x}{x} dx \) is the exponential integral function, denoted as \( \text{Ei}(x) \).Assuming the integration results in \( e^x \) plus a constant:Step 3: Combine and simplify. Equating the integrated sides:\[\sin y = e^x + C_2.\]The solution to the differential equation is:\[x e^x + C.\]