Solve :
dy/dx= 1/x+y+1
From the given question, we have;
"\\displaystyle\n\\frac{dy}{dx}=\\frac{1}{x+y+1}"
"\\displaystyle\n\\Rightarrow \\frac{dx}{dy}=\\frac{x+y+1}{1}=x+y+1\\\\\n\\Rightarrow \\frac{dx}{dy}-x=(y+1)\\qquad\\cdots\\cdots\\cdots\\cdots(1)"
which is a linear differential equation with x as dependent variable and y as independent variable.
Comparing (1) with the general form "\\displaystyle\n\\frac{dx}{dy}+Px=Q", which is a linear differential equation with x as dependent variable yields; "\\displaystyle\nP=-1,\\ \\text{and } Q=y+1."
Now, the integrating factor (I.F.) "\\displaystyle\n=e^{\\int Pdy}=e^{-\\int dy}=e^{-y}"
Multiplying (1) by I.F. yields;
"\\displaystyle\n\\frac{dx}{dy}e^{-y}-xe^{-y}=(y+1)e^{-y}\\\\\n\\Rightarrow \\frac{d(xe^{-y})}{dy}=(y+1)e^{-y}\\\\\n\\Rightarrow e^{-y}x=\\int(y+1)e^{-y}\\ dy=\u2212ye^{\u2212y}\u22122e^{\u2212y}+k", where k is an arbitrary constant.
"\\displaystyle\n\\Rightarrow e^{-y}x=-e^{-y}(y+2)+k\\\\\n\\Rightarrow x=ke^y-(y+2)"
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