+138 Differential equation\(\frac{dy}{dx}=1+\frac{y}{x}\)
Initial Condition (1,1)
My work as follows: First I rewrite as a first order linear
\(y'\:-\frac{1}{x}y=1\)
then find the integrating factor
\(μ\left(x\right)=\frac{1}{x}\)
Then multiply by integrating factor
\(\frac{y'\:}{x}-\frac{y}{x^2}=\frac{1}{x}\)
Then product rule
\(\left(\frac{1}{x}y\right)'\:=\frac{1}{x}\)
And then solve for this
\(y=\ln \left(x\right)x+cx\)
c=1,
so
\(y=\ln \left(x\right)x+x\)
Am I doing something wrong here? Any help is appreciated :)
+37146 Maybe you overthought it a bit, but I think it is just this: 
1 + y x-1 dy/dx
= -1 y x-2
= - y/x2 ??
I solved the de as you did and got exactly the same result.
You can check that it's correct by differentiating it and seeing that it takes you back to the original de.
So,
\(\displaystyle y = x \ln(x) +x, \\ y' = \ln(x) + x/x + 1 =\ln(x)+2, \\ xy'=x\ln(x)+2x =( x\ln(x)+x)+x=y+x,\\ y'=y/x+1,\)
which is the original de, so with you, I think that it's correct.
