Question

By substituting \[x\ =\ vy\]the transformed equation of \[\left( 1+{{e}^{\dfrac{x}{y}}} \right)dx+{{e}^{\dfrac{x}{y}}}\left( 1-\dfrac{x}{y} \right)dy\ =\ 0\] is

a)\[y\dfrac{dv}{dy}\ -\ \left( \dfrac{v+{{e}^{v}}}{1+{{e}^{v}}} \right)\ =\ 0\]
b)\[y\dfrac{dv}{dy}\ +\ \left( \dfrac{v+{{e}^{v}}}{1+{{e}^{v}}} \right)\ =\ 0\]
c)\[y\dfrac{dv}{dy}\ +x\ \left( v+{{e}^{v}} \right)\ =\ 0\]
d)\[y\dfrac{dy}{dv}\ +\ x\left( v+{{e}^{v}} \right)\ =\ 0\]

Answer 1

Hint: here in the given question the main point to observe is v is not a constant, v is also variable. So, when you substitute x as the given value into the given expression you must note that the value of v must be differentiated and the term $dv$ in all options will also support the same thing. So, the first thing you need to find is dx value in terms of $dv$ and $dy$ to substitute and make the given equation free of x term. It must only contain v and y terms. Even $dx$ must not be there in the final question. By doing all this you get a final result.

Complete step-by-step answer:
Given relation between x, v, y is as followed in the question:

\[x\ =\ vy\] …………………………………….(i)

By differentiating just d on both sides, we get it as:

\[dx\ =\ d\left( vy \right)\].

We know the $u-v$ rule by rules of basic differentiation:

\[d\left( uv \right)\ =\ vdu+udv\]

By substituting this in the above we get it as

\[dx\ =\ vdy+ydv\] ………………………………..(2)

Give differential equation in the question which we need:

\[\left( 1+{{e}^{\dfrac{x}{y}}} \right)dx+{{e}^{\dfrac{x}{y}}}\left( 1-\dfrac{x}{y} \right)dy\ =\ 0\] ……………………………..(3)

By substituting equation (i), equation (ii) into equation (iii), we get it as:

\[\left( 1+{{e}^{\dfrac{vy}{y}}} \right)\left( vdy+ydv \right)+{{e}^{\dfrac{vy}{y}}}\left( 1-v \right)dy\ =\ 0\]

By cancelling y in some terms, the equation gets the form:

 \[\left( 1+{{e}^{v}} \right)\left( vdy+ydv \right)+{{e}^{v}}\left( 1-v \right)dy\ =\ 0\]

By using the distributive law \[a\centerdot \left( b+c \right)\ =\ a\centerdot b+a\centerdot c\], we get:

\[vdy+ydv+v{{e}^{v}}dy+y{{e}^{v}}dv+{{e}^{v}}dy-{{e}^{v}}vdy\ =\ 0\]

By cancelling the terms and taking $dy$, $dv$ common, we get:

\[\left( v+{{e}^{v}} \right)dy+\left( y+y{{e}^{v}} \right)dv\ =\ 0\]

By sending the terms and rearranging them, the above equation turns into:

\[-\dfrac{ydv}{dy}\ =\ \left( \dfrac{v+{{e}^{v}}}{1+{{e}^{v}}} \right)\]

By rearranging final transformed: \[y\dfrac{dv}{dy}\ +\ \left( \dfrac{v+{{e}^{v}}}{1+{{e}^{v}}} \right)\ =\ 0\]

Therefore, option (b) is the correct answer for the given question.

Note: (1) Use the $u-v$ carefully as it defines the whole answer.

(2) Do the rearranging as per the options. Try to match with the form given.

(3) Final answer is not fixed; it is just one of all representations.