Solve the given differential equation by using an appropriate substitution. The DE is of the form dy dx = f(Ax + By + C)
dy/dx = (x + y + 1)^2
"\\frac{dy}{dx}=(x+y+1)\u00b2"
Let (x+y+1) = t
Differentiating with respect to x ,
1+ "\\frac{dy}{dx} = \\frac{dt}{dx}"
=> "\\frac{dy}{dx} = \\frac{dt}{dx}-1"
So the differential equation becomes
"\\frac{dt}{dx}-1 = t\u00b2"
=> "\\frac{dt}{dx} = (1+t\u00b2)"
=> "\\frac{dt}{1+t\u00b2} = dx"
Integrating ,
"\\int \\frac{dt}{1+t\u00b2} = \\int dx"
=> "tan^{-1}t = x +" C , where C is integration constant.
=> tan"^{-1}" ( x + y + 1 ) = x + C
So the solution of the given differential equation is tan"^{-1}" (x + y + 1) = x + C where C is integration constant.
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