Answer to Question #288876 in Differential Equations for Pankaj

Question

Solve :

dy/dx= 1/x+y+1

Answer 1

From the given question, we have;

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

$\displaystyle \Rightarrow \frac{dx}{dy}=\frac{x+y+1}{1}=x+y+1\\ \Rightarrow \frac{dx}{dy}-x=(y+1)\qquad\cdots\cdots\cdots\cdots(1)$

which is a linear differential equation with x as dependent variable and y as independent variable.

Comparing (1) with the general form $\displaystyle \frac{dx}{dy}+Px=Q$, which is a linear differential equation with x as dependent variable yields; $\displaystyle P=-1,\ \text{and } Q=y+1.$

Now, the integrating factor (I.F.) $\displaystyle =e^{\int Pdy}=e^{-\int dy}=e^{-y}$

Multiplying (1) by I.F. yields;

$\displaystyle \frac{dx}{dy}e^{-y}-xe^{-y}=(y+1)e^{-y}\\ \Rightarrow \frac{d(xe^{-y})}{dy}=(y+1)e^{-y}\\ \Rightarrow e^{-y}x=\int(y+1)e^{-y}\ dy=−ye^{−y}−2e^{−y}+k$, where k is an arbitrary constant.

$\displaystyle \Rightarrow e^{-y}x=-e^{-y}(y+2)+k\\ \Rightarrow x=ke^y-(y+2)$