The solution of the differential equation ≤ft(1+e^{\;x/y}) dx+e^{\;x/y} ≤ft(1+\;x/y) d y=0 is

Correct Answer is : ye xy+x=Cy e^{\;x/y} + x = Cyeyx+x=C

Correct Answer is : ye xy+x=Cy e^{\;x/y} + x = Cyeyx+x=C

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VITEEE 2025 Paper

Question : 88 of 125

Marks: +1, -0

\left(1+e^{\;\frac{x}{y}}\right) dx+e^{\;\frac{x}{y}} \left(1+\;\frac{x}{y}\right) d y=0

Solution:

The given differential equation is

\; \left(1+e^{\;\frac{x}{y}}\right) dx+e^{\;\frac{x}{y}} \left(1-\;\frac{x}{y}\right) d y=0

\;\; \frac{dx}{dy} = \frac{e^{\frac{x}{y}} \left(\frac{x}{y}-1\right)}{e^{x/y}+1}

\; = g\!\left(\;\frac{x}{y}\right) \because \frac{dx}{dy}=g\!\left(\;\frac{x}{y}\right)

∴ eq. (i) is the homogeneous differential equation so, put

\frac{x}{y}=v \; \text{ i.e., } \; x=v y \Rightarrow \frac{dx}{dy}=v+y \frac{dv}{dy}

Then, eq. (i) becomes

v+y \frac{dv}{dy} = \frac{e^v(v-1)}{e^v+1} \Rightarrow y \frac{dv}{dy} = \frac{e^v(v-1)}{e^v+1} - v

\Rightarrow y \frac{dv}{dy} = \frac{v^v - e^v - v e^v - v}{e^v+1} - v

\Rightarrow \; \frac{e^v+1}{e^v+v} dv = -\frac{1}{y} dy

On integrating both sides, we get

\int \frac{e^v+1}{e^v+v} dv = - \int \frac{1}{y} dy

\Rightarrow e^v+1 = \frac{dt}{dv} \Rightarrow dv = \frac{dt}{e^v+1}

\therefore \int \frac{e^v+1}{t} \frac{dt}{e^v+1} - \log |y| + \log C

\Rightarrow \log |t| + \log |y| = \log C

\Rightarrow \log |e^v+v| + \log |y| = \log C

\Rightarrow \log |(e^v+v) y| = C \Rightarrow |(e^v+v) y| = C

\Rightarrow (e^v+v) y = C . \; \text{ So, put } \; v = \frac{x}{y}, \; \text{ we get } \;

\; \left(e^{x/y} + \frac{x}{y}\right) t = e^v+v)

\;

This is the required solution of the given differential equation.

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