# Solve: $(x+y)(dx-dy)=dx+dy.$[Hint:Substitute x+y=z after separating dx and dy]

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

Toolbox:
• A linear differential equation of the form.$\large\frac{dy}{dx}$$=(x+y) can be solved by substituting x+y=z and \large\frac{dy}{dx}=\frac{dz}{dx}$$-1$
• Then the variable can be seperated and integrated.
• $\int \large\frac{dx}{x}$$=\log x Given (x+y)(dx-dy)=dx+dy This can be written as xdx-xdy+ydx-ydy=dx+dy => dx(x+y-1)=dy(x+y+1) (ie)dx(x+y-1)=dy(x+y+1) =>\large\frac{dy}{dx}=\large\frac{x+y-1}{x+y+1}-----(1) Let x+y=z on differentiating w.r.t x we get 1+\large\frac{dy}{dx}=\frac{dz}{dx} or \large\frac{dy}{dx}=\frac{dz}{dx}-1 Now substituting in equ (1) we get, \large\frac{dz}{dx}$$-1=\large\frac{z-1}{z+1}$
=>$\large\frac{dz}{dx}=\frac{z-1}{z+1}$$+1 =\large\frac{z-1+z+1}{z-1} \large\frac{dz}{dx}=\frac{2z}{z-1} On seperating the variables we get \bigg(\large\frac{z-1}{2z}\bigg)$$dz=dx$
=>$\large\frac{1}{2}\bigg(\frac{z-1}{z}\bigg)$$dz=dx On integrating we get \large\frac{1}{2}\bigg[\int \frac{z}{z}$$dz-\int \large\frac{1}{z}$$dz\bigg]=\int dx =>\large\frac{1}{2}\bigg[\int$$dz-\int \large\frac{1}{z}$$dz\bigg]=\int dx \large\frac{1}{2}\bigg[$$z-\log z\bigg]=x+c$
$z-\log z=2x+c$
Now substituting for $z=(x+y)$
$x+y-\log(x+y)=2x+c$
$y=x+\log (x+y)+c$
or $y-x=\log _e (x+y)$
$c\;e^{\large y-x}=x+y$