Differential equations in function Equations (1) : $xy'+(1-x)y=1$  let $z=xy+1$

determine and solve the differential equation (2) whose general solution is the function $z$ 
.

-determine the general solution of (1)
If $z=xy+1$ then
\begin{eqnarray}
\frac{dz}{dx} &amp;=&amp; \frac{d}{dx}\left(xy+1\right) \\
              &amp;=&amp; y+x\frac{dy}{dx}
\end{eqnarray}
Now:
\begin{eqnarray}
x\frac{dy}{dx}+y-xy       &amp;=&amp; 1 \\
\implies x\frac{dy}{dx}+y &amp;=&amp; 1+xy\\
\implies \frac{dz}{dx}   &amp;=&amp; z
\end{eqnarray}
Now you solve this equation for $z$ and then change back the variables to $x$ and $y$ at the end.
Compute $z'=y+xy'=y+1-(1-x)y=z$

So the general solution of the above equation (call it eqn (2))is

$$z(x)=z(0)e^x=e^x$$

because $z(0)=1$. Solving then for $y$:

$$y(x)=\frac{z-1}{x}=\frac{e^x-1}{x}$$