## anonymous one year ago all lines whose slopes and x intecept equal Ans. y=y'x-y'^2

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1. anonymous

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 $$C=m$$.

2. anonymous

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