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

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

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

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