(1 + e^{y/x})dy + e^{y/x}(1 – \frac{y}{x}) = 0

Question 1:

Solve the following differential equation: (1 + e^{y/x})dy + e^{y/x}(1 - \frac{y}{x}) = 0 x ≠ 0

Solution:

(1 + e^{y/x})dy + e^{y/x}(1 - \frac{y}{x}) = 0

\Rightarrow (1 + e^{y/x}) = (\frac{y}{x} -1)e^{y/x} dx

\Rightarrow \dfrac{dy}{dx} = \dfrac{(\frac{y}{x}-1)e^{y/x}}{(1 + e^{y/x}}

It is a homogeneous differential equation.

Put y = vx and \frac{dy}{dx} = v + x \frac{dv}{dx}

We have,

\displaystyle v + x \frac{d v}{dx} = \frac{(v - 1)e^v}{1 + e^v}

\displaystyle \Rightarrow v + x \frac{d v}{dx} = \frac{(e^v v - e^v)}{1 + e^v}

\displaystyle \Rightarrow x \frac{d v}{dx} = \frac{(e^v v - e^v)}{1 + e^v} - v

\displaystyle\Rightarrow x \frac{d v}{dx} = \frac{v e^v - e^v - v - v e^v}{1 + e^v}

\displaystyle\Rightarrow x \frac{d v}{dx} = - \frac{v + e^v}{1 + e^v}

\displaystyle\Rightarrow \dfrac{1 + e^v}{v + e^v} dv = -\dfrac{dx}{x}  ………(i)

On integrating both sides, we have

\displaystyle \int\frac{1 + e^v}{v + e^v} dv = - \int \frac{dx}{x}

Let \displaystyle v + e^v = t \Rightarrow (1 + e^v) dv = dt

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

\Rightarrow \displaystyle \int\frac{1 + e^v}{t} \frac{dv}{1 + e^v} = - \int \frac{dx}{x}

\Rightarrow \displaystyle \int \frac{dt}{t} = -\log |x| + \log |C|

\Rightarrow \displaystyle \log |t| + \log |x| = \log |C|

\Rightarrow \displaystyle \log |v + e^v| + \log |x| = \log |C|

\Rightarrow \displaystyle \log |x(v + e^v}| = \log |C|

\displaystyle \Rightarrow x(v + e^v) = C

\displaystyle \Rightarrow x(\frac{y}{x} + e^{y/x}) = C

\displaystyle \Rightarrow y + x e^{y/x} = C

Team Gmath

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