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: $e^{xy-lnx^2} = e^{xy} \cdot e^{-lnx^2} = e^{xy} \cdot \frac{1}{x^2}$ Substitute this back into the equation: $\left(x \frac{dy}{dx}+y\right) = \frac{e^{xy}}{x^2} \left(x \frac{dy}{dx}-y\right)$ Multiply both sides by $x^2$: $x^2 \left(x \frac{dy}{dx}+y\right) = e^{xy} \left(x \frac{dy}{dx}-y\right)$ Rearrange the terms to group $\frac{dy}{dx}$ and $y$: $x^3 \frac{dy}{dx} + x^2 y = x e^{xy} \frac{dy}{dx} - y e^{xy}$ $x^3 \frac{dy}{dx} - x e^{xy} \frac{dy}{dx} = -x^2 y - y e^{xy}$ $\frac{dy}{dx} (x^3 - x e^{xy}) = -y (x^2 + e^{xy})$ $\frac{dy}{dx} = \frac{-y (x^2 + e^{xy})}{x (x^2 - e^{xy})}$ This does not seem to lead to a simple separation. Let's try another manipulation.

Divide the original equation by $x^2$: $\frac{x \frac{dy}{dx}+y}{x^2} = \frac{e^{xy-lnx^2}}{x^2} \left(x \frac{dy}{dx}-y\right)$ Recognize that $\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{x \frac{dy}{dx}-y}{x^2}$ and $\frac{d}{dx}(xy) = x \frac{dy}{dx}+y$. So the equation becomes: $\frac{1}{x^2} \frac{d}{dx}(xy) = \frac{e^{xy}}{x^4} \left(x \frac{dy}{dx}-y\right)$ This is still not directly separable.

Let's divide the original equation by $x^2 e^{xy}$: $\frac{x \frac{dy}{dx}+y}{x^2 e^{xy}} = \frac{e^{xy-lnx^2}}{x^2 e^{xy}} \left(x \frac{dy}{dx}-y\right)$ $\frac{x \frac{dy}{dx}+y}{x^2 e^{xy}} = \frac{1}{x^2 e^{xy}} \left(x \frac{dy}{dx}-y\right)$ This is not simplifying.

Let's rearrange the equation as: $\left(x \frac{dy}{dx}+y\right) = \frac{e^{xy}}{x^2} \left(x \frac{dy}{dx}-y\right)$ Divide both sides by $x^2$: $\frac{x \frac{dy}{dx}+y}{x^2} = \frac{e^{xy}}{x^4} \left(x \frac{dy}{dx}-y\right)$ $\frac{1}{x^2}\frac{d}{dx}(xy) = \frac{e^{xy}}{x^2} \frac{1}{x^2}(x \frac{dy}{dx}-y)$ $\frac{1}{x^2}\frac{d}{dx}(xy) = \frac{e^{xy}}{x^2} \frac{d}{dx}\left(\frac{y}{x}\right)$ This is still not separable.

Let's try dividing the original equation by $x^2$: $\frac{x \frac{dy}{dx}+y}{x^2} = \frac{e^{xy}}{x^4} \left(x \frac{dy}{dx}-y\right)$ $\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{e^{xy}}{x^2} \frac{1}{x^2} \left(x \frac{dy}{dx}-y\right)$ $\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{e^{xy}}{x^2} \frac{d}{dx}\left(\frac{y}{x}\right)$ This equation implies either $\frac{d}{dx}\left(\frac{y}{x}\right) = 0$ or $1 = \frac{e^{xy}}{x^2}$.

Case 1: $\frac{d}{dx}\left(\frac{y}{x}\right) = 0$ This means $\frac{y}{x} = c$, or $y = cx$. Substituting this into the original equation gives $x(c) + cx = e^{cx^2 - \ln x^2} (x(c) - cx)$, which simplifies to $2cx = 0$. This implies $c=0$, so $y=0$, which is a trivial solution.

Case 2: $1 = \frac{e^{xy}}{x^2}$ $e^{xy} = x^2$ Taking the natural logarithm on both sides: $xy = \ln(x^2)$ $xy = 2 \ln x$ This is not in the form of the options.

Let's go back to the equation: $\left(x \frac{dy}{dx}+y\right) = \frac{e^{xy}}{x^2} \left(x \frac{dy}{dx}-y\right)$ Divide by $x^2$: $\frac{x \frac{dy}{dx}+y}{x^2} = \frac{e^{xy}}{x^4} \left(x \frac{dy}{dx}-y\right)$ $\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{e^{xy}}{x^2} \frac{1}{x^2} \left(x \frac{dy}{dx}-y\right)$ $\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{e^{xy}}{x^2} \frac{d}{dx}\left(\frac{y}{x}\right)$ This is incorrect.

Let's re-examine the original equation: $\left(x \frac{dy}{dx}+y\right) = e^{xy-lnx^2} \left(x \frac{dy}{dx}-y\right)$ $\frac{x \frac{dy}{dx}+y}{x \frac{dy}{dx}-y} = e^{xy-lnx^2}$ $\frac{x \frac{dy}{dx}+y}{x \frac{dy}{dx}-y} = \frac{e^{xy}}{x^2}$ $x^2 \left(x \frac{dy}{dx}+y\right) = e^{xy} \left(x \frac{dy}{dx}-y\right)$ $x^3 \frac{dy}{dx} + x^2 y = x e^{xy} \frac{dy}{dx} - y e^{xy}$ $x^3 \frac{dy}{dx} - x e^{xy} \frac{dy}{dx} = -x^2 y - y e^{xy}$ $\frac{dy}{dx} (x^3 - x e^{xy}) = -y (x^2 + e^{xy})$ $\frac{dy}{dx} = \frac{-y (x^2 + e^{xy})}{x (x^2 - e^{xy})}$

Let's try dividing the original equation by $x^2$: $\frac{x \frac{dy}{dx}+y}{x^2} = \frac{e^{xy}}{x^4} \left(x \frac{dy}{dx}-y\right)$ $\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{e^{xy}}{x^2} \frac{1}{x^2} \left(x \frac{dy}{dx}-y\right)$ $\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{e^{xy}}{x^2} \frac{d}{dx}\left(\frac{y}{x}\right)$ This is incorrect.

Let's divide the original equation by $x^2$: $\frac{x \frac{dy}{dx}+y}{x^2} = \frac{e^{xy-lnx^2}}{x^2} \left(x \frac{dy}{dx}-y\right)$ $\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{e^{xy}}{x^4} \left(x \frac{dy}{dx}-y\right)$ This is incorrect.

Let's divide the original equation by $x^2$: $\frac{x \frac{dy}{dx}+y}{x^2} = \frac{e^{xy}}{x^4} \left(x \frac{dy}{dx}-y\right)$ $\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{e^{xy}}{x^2} \frac{1}{x^2} (x \frac{dy}{dx}-y)$ $\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{e^{xy}}{x^2} \frac{d}{dx}\left(\frac{y}{x}\right)$ This is incorrect.

Let's rewrite the equation as: $\frac{x \frac{dy}{dx}+y}{e^{xy}} = \frac{1}{x^2} \left(x \frac{dy}{dx}-y\right)$ Multiply by $dx$: $\frac{x dy + y dx}{e^{xy}} = \frac{x dy - y dx}{x^2}$ $\frac{d(xy)}{e^{xy}} = d\left(\frac{y}{x}\right)$ Integrate both sides: $\int \frac{d(xy)}{e^{xy}} = \int d\left(\frac{y}{x}\right)$ Let $u = xy$, then $du = (x dy + y dx)$. $\int \frac{du}{e^u} = \frac{y}{x} + c$ $\int e^{-u} du = \frac{y}{x} + c$ $-e^{-u} = \frac{y}{x} + c$ Substitute back $u = xy$: $-e^{-xy} = \frac{y}{x} + c$ $\frac{y}{x} + e^{-xy} = -c$ Let $C = -c$. $\frac{y}{x} + e^{-xy} = C$ This matches the third option.