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# Solve the differential equation $\Large\frac{dy}{dx}\normalsize+1=e^{x+y}$

$\begin{array}{1 1}(A)\;(x+c)e^{(\large x-y)}+1=0 \\ (B)\;(x+c)e^{(\large x+y)}-1=0 \\(C)\;(x+c)e^{(\large x+y)}+1=0 \\(D)\;(x-c)e^{(\large x+y)}-1=0\end{array}$

Toolbox:
• A linear differential equation is of the form $\large\frac{dy}{dx}$$=f(x+y), then (x+y) can be substituted as z and \large\frac{dy}{dx}=\frac{dz}{dx}-1.Then the variable can be seperated and integrated, to get the solution. • \int e^{-x}dx=-e^{-x}+c Given \large\frac{dy}{dx}$$+1=e^{x+y}$-----(1)
Put $x+y=z$ on differentiating w.r.t x we get,
$1+\large\frac{dy}{dx}=\frac{dz}{dx}$$=>\large\frac{dy}{dx}=\frac{dz}{dx}$$-1$
Now substituting in equ(1) we get
$\large\frac{dz}{dx}$$-1+1=e^z => \large\frac{dz}{dx}$$=e^z$
Now seperating the variables we get
$\Large\frac{dz}{e^z}$$=dx =>e^{-z}dz=dx On integrating we get: \int e^{-z}dz=\int dx -e^{-z}=x+c Substituting for z we get -e^{-(x+y)}=x+c Dividing throughout by e^{-(x+y)} we get -1=e^{(x+y)}(x+c) Therefore (x+c)e^{(\large x+y)}$$+1=0$

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