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(C) Given differential equation: $(e^{y-x}) dy = (e^x - e^y) dx$ Multiply both sides by $e^x$: $e^y dy = (e^{2x} - e^x e^y) dx$ Rearrange the terms: $

Vedclass · Accepted Answer

$e^y e^{e^x} = e^x e^{e^x} - e^{e^x} + c$

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(C) Given differential equation: $(e^{y-x}) dy = (e^x - e^y) dx$ Multiply both sides by $e^x$: $e^y dy = (e^{2x} - e^x e^y) dx$ Rearrange the terms: $e^y \frac{dy}{dx} + e^x e^y = e^{2x}$ Let $z = e^y$,then $\frac{dz}{dx} = e^y \frac{dy}{dx}$. The equation becomes: $\frac{dz}{dx} + e^x z = e^{2x}$. This is a linear differential equation of the form $\frac{dz}{dx} + P(x)z = Q(x)$,where $P(x) = e^x$ and $Q(x) = e^{2x}$. Integrating factor $IF = e^{\int P(x) dx} = e^{\int e^x dx} = e^{e^x}$. The solution is $z \cdot IF = \int Q(x) \cdot IF dx + c$. $z \cdot e^{e^x} = \int e^{2x} \cdot e^{e^x} dx + c$. Let $u = e^x$,then $du = e^x dx$. $z \cdot e^{e^x} = \int u \cdot e^u du + c$. Using integration by parts: $\int u e^u du = u e^u - e^u + c$. Substituting back $z = e^y$ and $u = e^x$: $e^y e^{e^x} = e^x e^{e^x} - e^{e^x} + c$.

Find the solution of the differential equation $(e^{y-x}) dy = (e^x - e^y) dx$.

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