Question

What will be the transformed equation of $\left( {1 + {e^{\dfrac{x}{y}}}} \right)dx + {e^{\dfrac{x}{y}}}\left( {1 - \dfrac{x}{y}} \right)dy = 0$ by making the substitution of x=vy
$(a){\text{ }}y\dfrac{{dv}}{{dy}} - \dfrac{{\left( {1 + {e^v}} \right)}}{{\left( {1 - {e^v}} \right)}} = 0 \\\ (b){\text{ }}y\dfrac{{dv}}{{dy}} + \dfrac{{\left( {v + {e^v}} \right)}}{{\left( {1 + {e^v}} \right)}} = 0 \\\ (c){\text{ }}y\dfrac{{dv}}{{dy}} + x\left( {v + {e^v}} \right) = 0 \\\ (d){\text{ }}y\dfrac{{dy}}{{dv}} + x\left( {v + {e^v}} \right) = 0 \\\$

Answer 1

Hint – Differentiate x=vy both the sides with respect to x to obtain the value of dx. Put this value of dx into the main equation and then simplify. This will help getting the answer.

Complete step-by-step answer:

Given equation is
$\left( {1 + {e^{\dfrac{x}{y}}}} \right)dx + {e^{\dfrac{x}{y}}}\left( {1 - \dfrac{x}{y}} \right)dy = 0$………………. (1)

Now we have to substitute $x = vy$ in the given equation.

Now differentiate this equation w.r.t. x according to the product rule of differentiation we have,
$\Rightarrow 1 = v\dfrac{{dy}}{{dx}} + y\dfrac{{dv}}{{dx}}$

Now above equation is also written as
$\Rightarrow dx = vdy + ydv$

Now substitute the values of x and dx in equation (1) we have,
$\Rightarrow \left( {1 + {e^{\dfrac{{vy}}{y}}}} \right)\left( {vdy + ydv} \right) + {e^{\dfrac{{vy}}{y}}}\left( {1 - \dfrac{{vy}}{y}} \right)dy = 0$

Now simplify the above equation we get,
$\Rightarrow \left( {1 + {e^v}} \right)\left( {vdy + ydv} \right) + {e^v}\left( {1 - v} \right)dy = 0$
$\Rightarrow vdy + ydv + v{e^v}dy + y{e^v}dv + {e^v}dy - v{e^v}dy = 0$

Now cancel out the terms and the remaining terms are,
$\Rightarrow vdy + ydv + y{e^v}dv + {e^v}dy = 0$
Now take common (ydv) from second and third term and (dy) from first and fourth term we have,
$\Rightarrow ydv\left( {1 + {e^v}} \right) + dy\left( {v + {e^v}} \right) = 0$
$\Rightarrow ydv\left( {1 + {e^v}} \right) = - dy\left( {v + {e^v}} \right)$

Now divide by (dy) we have,
$\Rightarrow y\dfrac{{dv}}{{dy}}\left( {1 + {e^v}} \right) = - \left( {v + {e^v}} \right)$
$\Rightarrow y\dfrac{{dv}}{{dy}} = - \dfrac{{\left( {v + {e^v}} \right)}}{{\left( {1 + {e^v}} \right)}}$
$\Rightarrow y\dfrac{{dv}}{{dy}} + \dfrac{{\left( {v + {e^v}} \right)}}{{\left( {1 + {e^v}} \right)}} = 0$

Hence option (B) is correct.

Note – The given equation is a differential equation, a differential equation is on that relates one or more functions and their derivatives. Direct substitution of x=vy without differentiating wouldn’t have helped getting anywhere so this was the tricky concept of this problem statement.