The general solution of ≤ft(1+e^{x/y}) dx + e^{x/y} ≤ft(1-x/y) dy = 0 is

Correct Answer is : yexyy e^{x/y}yeyx + x = c

Correct Answer is : yexyy e^{x/y}yeyx + x = c

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AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

Section: Mathematics

Question : 80 of 160

Marks: +1, -0

The general solution of

\left(1+e^{\frac{x}{y}}\right)

e^{\frac{x}{y}} \left(1-\frac{x}{y}\right)

dy = 0 is

Solution:

It is given,

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

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

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

......(i)

Substituting

x=vy

and differentiating it with respect to

y

\frac{dx}{dy}=v \frac{dy}{dy}+y \frac{dv}{dy}

\frac{dx}{dy}=v+y \frac{dv}{dy}

Substitute these values in equation (i).

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

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

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

-\frac{dy}{y}=\left[ \frac{1+e^{v}}{v+e^{v}} \right] dv

Integrating both sides we get,

-\int \frac{dy}{y}=\int \left[ \frac{1+e^{x}}{v+e^{v}} \right] dv

\int \frac{1+e^{x}}{v+e^{v}} dv=-\log y+\log c

Put,

v+e^{v}=t

(1+e^{v}) dv=dt

\therefore \int \frac{dt}{t}=-\log y+\log c

\log t=-\log y+\log c

\log y (v+e^{v})=\log c

Put

v

\frac{x}{y}

in above.

\log y \left( \frac{x}{y}+e^{\frac{x}{y}} \right)=\log c

y \left( \frac{x}{y}+e^{\frac{x}{y}} \right)=c

x+y e^{\frac{x}{y}}=c

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