Question

all lines whose slopes and x intecept equal Ans. y=y'x-y'^2

Answer 1

Any linear function takes the form \(y=mx+b\), where \(m\) is the slope and \(b\) is the intercept. Fix \(m=b\), then take your derivatives.
\[y=m(x+1)~~\implies~~y'=m\]
Back-substituting, you see that
\[y=y'(x+1)=y'x+y'\]
will work. I don't see a reason to have \(-(y')^2\) there... Just to check that this is indeed a sufficient ODE, let's solve it:
\[y=y'x+y'~~\iff~~y=(x+1)\frac{dy}{dx}~~\implies~~\frac{dy}{y}=\frac{dx}{x+1}\]
Integrating yields
\[\ln |y|=\ln|x+1|+C=\ln|x+1|+\ln|C|=\ln|C(x+1)|~~\implies~~y=C(x+1)\]
where \(C=m\).

Answer 2

The given solution seems more along the lines of the ODE whose solutions are all lines \(y=mx+b\) with \(b=-m^2\).