Question

By substituting $x=vy$ the transformed equation of $(1+e^{x/y})dx+e^{x/y}(1-\frac{x}{y})dy=0$ is

• A. $y\frac{dv}{dy}-\left(\frac{1+e^v}{1-e^v}\right)=0$
• B. $y\frac{dy}{dv}+\left(\frac{v+e^v}{1+e^v}\right)=0$
• C. $y\frac{dv}{dy}+x(v+e^v)=0$
• D. $y\frac{dy}{dv}+x(v+e^v)=0$

Answer 1

The correct option is B $y\frac{dy}{dv}+\left(\frac{v+e^v}{1+e^v}\right)=0$

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

$x=vy$

$\Rightarrow x/y=v$

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

From $\left(i\right)$

$(1+e^v)\frac{dx}{dy}+e^v(1-v)=0$

$\Rightarrow (1+e^v)(v+y\frac{dv}{dy})+e^v(1-v)=0$

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

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

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

$\Rightarrow y\frac{dv}{dy}+\frac{(v+e^v)}{1+e^v}=0$