Question

Solution of the differential equation $\frac{dy}{dx}+1=e^{x+y}$ is

• A. $e^{-(x+y)}+y=c$
• B. $e^{x+y}+x+y=0$
• C. $e^{-(x+y)}+x+c=0$
• D. $e^{-(x+y)}+y+2x=c$

Answer 1

Answer: (C)

$e^{-(x+y)}+x+c=0$

$\frac{dy}{dx}+1=e^{x+y}$

$\Rightarrow \frac{dy}{dx}+1=e^x.e^y$

$\Rightarrow e^{-y}\frac{dy}{dx}+e^{-y}=e^x$

putting $e^{-y}=t$

$-e^{-y}\frac{dy}{dx}=\frac{dt}{dx}$

$\Rightarrow -\frac{dy}{dx}+t=e^x$

$\Rightarrow \frac{dy}{dx}-t=-e^x$

here $P=-1Q=-e^x$

$I.F=e^{\int Pdx}=e^{\int -1dx}=-e^x$

So , solution of D.E is

$t\times I.F=\int Q\times I.Fdx+c$

$\Rightarrow t\times e^{-x}=\int -e^x\times e^{-x}dx+c$

$\Rightarrow e^{-y}{e}^{-x}=\int -1dx+c$

$\Rightarrow e^{-\left(x+y\right)}=-x+c$

$\Rightarrow e^{-\left(x+y\right)}+x=c$

$\Rightarrow e^{-\left(x+y\right)}+x+c=0$