The general solution of the differential equa | vedclass

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Question

(C) Given the differential equation: $\left(1+e^{\frac{x}{y}}\right) dx + \left(1-\frac{x}{y}\right) e^{\frac{x}{y}} dy = 0$ Rearranging the terms: $\

Vedclass · Accepted Answer

$x+y e^{\frac{x}{y}}=C$

Vedclass · Answer

(C) Given the differential equation: $\left(1+e^{\frac{x}{y}}\right) dx + \left(1-\frac{x}{y}\right) e^{\frac{x}{y}} dy = 0$ Rearranging the terms: $\frac{dx}{dy} = \frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}$ ...$(i)$ This is a homogeneous differential equation. Let $x = vy$, then $\frac{dx}{dy} = v + y \frac{dv}{dy}$ ...(ii) Substituting (ii) into $(i)$: $v + y \frac{dv}{dy} = \frac{-e^v(1-v)}{1+e^v}$ $y \frac{dv}{dy} = \frac{-e^v + ve^v}{1+e^v} - v = \frac{-e^v + ve^v - v - ve^v}{1+e^v} = \frac{-(e^v + v)}{1+e^v}$ Separating variables: $\frac{1+e^v}{v+e^v} dv = -\frac{dy}{y}$ Integrating both sides: $\int \frac{1+e^v}{v+e^v} dv = -\int \frac{dy}{y}$ Let $v+e^v = t$, then $(1+e^v) dv = dt$. So, $\ln|t| = -\ln|y| + \ln|C|$ $\ln|v+e^v| + \ln|y| = \ln|C| \Rightarrow y(v+e^v) = C$ Substituting $v = \frac{x}{y}$: $y(\frac{x}{y} + e^{\frac{x}{y}}) = C \Rightarrow x + ye^{\frac{x}{y}} = C$

The general solution of the differential equation $\left(1+e^{\frac{x}{y}}\right) dx + \left(1-\frac{x}{y}\right) e^{\frac{x}{y}} dy = 0$ is ($C$ is an arbitrary constant).

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