Question

Solution of the differential equation $\left(\frac{dy}{dx}\right)=e^{x-y}(e^x-e^y)$ is:
(where $c$ is integration constant)

• A. $e^y{e}^{e^x}=e^x{e}^{e^x}+c$
• B. $e^y{e}^{e^x}=e^{e^x}+c$
• C. $e^y{e}^x=e^x{e}^{e^x}-e^{e^x}+c$
• D. $e^y{e}^{e^x}=e^x{e}^{e^x}-e^{e^x}+c$

Answer 1

The correct option is D $e^y{e}^{e^x}=e^x{e}^{e^x}-e^{e^x}+c$

Multiplying the given equation by $e^y$, we get
$e^y\frac{dy}{dx}+e^x{e}^y=e^{2x}\cdots \left(1\right)$
Putting $e^y=v,$ so that $e^y\frac{dy}{dx}=\frac{dv}{dx},$
and equation $\left(1\right)$ transform to $\frac{dv}{dx}+e^{x}v=e^{2x}$
On comparing with $\frac{dv}{dx}+Pv=Q,$
We get $P=e^x$ and $Q=e^{2x}$
$I.F.=e^{\int e^{x}dx}=e^{e^x}$

Hence solution is $v{e}^{e^x}=\int e^{2x}{e}^{e^x}dx$
$\Rightarrow v{e}^{e^x}=\int \left(e^x\right)e^{\left(e^x\right)}\left(e^{x}dx\right)$
Let $e^x=t\Rightarrow e^{x}dx=dt$
$\Rightarrow v{e}^{e^x}=\int t{e}^{t}dt$
$\Rightarrow e^y{e}^{e^x}=t{e}^t-e^t+c$
$\Rightarrow e^y{e}^{e^x}=e^x{e}^{e^x}-e^{e^x}+c$