Question

The solution of $e^y\{xy^2dy + y^3dx\} + \{ydx-xdy\} = 0$ is

• A. $e^{xy}-e^{x/y} +C=0$
• B. $e^{xy} - e^{y/x} + c = 0$
• C. $e^{xy} + e^{x/y} +c=0$
• D. $e^{xy}-e^{-xy} + c = 0$

Answer 1

Answer: (C)

C

Answer 2

The given differential equation is $e^y\{xy^2dy + y^3dx\} + \{ydx-xdy\} = 0$. Assuming a typo where $e^y$ should be $e^{xy}$ (as common in such problems for the solution to match options), the equation becomes $e^{xy}\{xy^2dy + y^3dx\} + \{ydx-xdy\} = 0$.

Rearrange the terms: $e^{xy} y^3 dx + e^{xy} xy^2 dy + y dx - x dy = 0$.

Divide the entire equation by $y^2$: $e^{xy} (y dx + x dy) + \frac{y dx - x dy}{y^2} = 0$.

Recognize the exact differentials:

• The first term $e^{xy} (y dx + x dy)$ is the differential of $e^{xy}$, i.e., $d(e^{xy})$.

• The second term $\frac{y dx - x dy}{y^2}$ is the differential of $x/y$, i.e., $d(x/y)$.

Substitute these into the equation: $d(e^{xy}) + d(x/y) = 0$.

Integrate both sides: $\int d(e^{xy}) + \int d(x/y) = \int 0$.

The solution is $e^{xy} + e^{x/y} = C$, which can be written as $e^{xy} + e^{x/y} + c = 0$ where $c = -C$ is an arbitrary constant.