Question

Solution of the differential equation $\left(x\frac{dy}{dx}+y\right)=e^{xy-lnx^2}\left(x\frac{dy}{dx}-y\right)$:

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

Answer 1

Answer: (C)

$\frac{y}{x}+e^{-xy}=c$

Answer 2

The given differential equation is: $\left(x\frac{dy}{dx}+y\right)=e^{xy-lnx^2}\left(x\frac{dy}{dx}-y\right)$ First, simplify the exponential term: $e^{xy-lnx^2} = e^{xy} \cdot e^{-lnx^2} = e^{xy} \cdot e^{ln(x^{-2})} = e^{xy} \cdot x^{-2} = \frac{e^{xy}}{x^2}$ Substitute this back into the differential equation: $x\frac{dy}{dx}+y = \frac{e^{xy}}{x^2}\left(x\frac{dy}{dx}-y\right)$ We can recognize that the left side is the derivative of the product $xy$: $\frac{d}{dx}(xy) = x\frac{dy}{dx}+y$ And the term $x\frac{dy}{dx}-y$ is related to the derivative of the quotient $\frac{y}{x}$: $\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{x\frac{dy}{dx}-y}{x^2} \implies x\frac{dy}{dx}-y = x^2\frac{d}{dx}\left(\frac{y}{x}\right)$ Substitute these into the differential equation: $\frac{d}{dx}(xy) = \frac{e^{xy}}{x^2}\left(x^2\frac{d}{dx}\left(\frac{y}{x}\right)\right)$ $\frac{d}{dx}(xy) = e^{xy}\frac{d}{dx}\left(\frac{y}{x}\right)$ Let $u = xy$ and $v = \frac{y}{x}$. The equation becomes: $\frac{du}{dx} = e^u \frac{dv}{dx}$ This is a separable differential equation. Rearrange and integrate: $\frac{1}{e^u} du = dv$ $e^{-u} du = dv$ $\int e^{-u} du = \int dv$ $-e^{-u} = v + C$ Substitute back $u = xy$ and $v = \frac{y}{x}$: $-e^{-xy} = \frac{y}{x} + C$ Rearranging the terms: $\frac{y}{x} + e^{-xy} = -C$ Let $c = -C$ be a new arbitrary constant. $\frac{y}{x} + e^{-xy} = c$