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Find out the differential equation of the following two families of curves :

  1. Straight lines having slope and $x$-intercept equal in magnitude.

  2. Straight lines at a fixed distance $p$ from the origin.

My Approach :

  1. a straight line is defined by $y = mx + c$, $x$-intercept $= -c/m$ but $-c/m = m$, so $c = -m^2$. $y = mx - m^2\; , \; dy/dx = m\; , \; y' = m$
  2. Straight lines at a fixed distance $p$ from the origin : $x\cos A + y\sin A = p,$ $y\sin A = p - x \cos A,$ $y = p\sin A - x\cot A,$ $y' = -\cot A.$

Are my answers correct?

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    @GerryMyerson I would accept $y=xy'-(y')^2$ as a correct answer, but I don't think you would since $y=x^2/4$ is also a solution of that DE, and according to you one should provide a DE satisfied only by $y=mx-m^2$!2012-07-01

5 Answers 5

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Here is an attempt to find a differential equation whose solutions are precisely the lines at distance $p$ from the origin.

We follow the question as far as $y'=-\cot A$ and then we try to eliminate $A$ in favor of $x,y,p$. We go back to $x\cos A+y\sin A=p$ Dividing through by $\sqrt{x^2+y^2}$, letting $\theta=\arctan(y/x)$, and using $\cos(r-s)=\cos r\cos s+\sin r\sin s$ we get $\cos(A-\theta)={p\over\sqrt{x^2+y^2}}$ Solving for $A$ we get $A=\theta+\arccos{p\over\sqrt{x^2+y^2}}$ so we have the differential equation $y'=-\cot\left(\arctan(y/x)+{p\over\sqrt{x^2+y^2}}\right)$ I managed to "simplify" this to $y'={y\tan{p\over\sqrt{x^2+y^2}}-x\over y+x\tan{p\over\sqrt{x^2+y^2}}}$

There must be a better way.

EDIT: Maybe it looks a little better as $xy'+y(1+(y')^2)=p\sqrt{1+(y')^2}$ or maybe not.

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    @MrR, I might have had a good answer for that in 2012, but I'm not seeing it now. All I can see is $xy'=-x\cot A=-x\cos A/\sin A=-(p-\sin A)/\sin A=-p\csc A+y=-p\sqrt{1+(y')^2}+y$, which is not what I wrote in my edit.2017-12-17
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Mercy and Gerry, I am solving the DE that I created .

$2\frac{dy}{dx} - x = \sqrt {x^2 - 4y}$

Let $ z^2 = ( {x^2 - 4y})$ $ 2z\frac{dz}{dx} = 2x -4\frac{dy}{dx}.$ $ z\frac{dz}{dx} = x -2\frac{dy}{dx}.$ Now $2\frac{dy}{dx} - x = \sqrt {x^2 - 4y}=z$ Simplifying we get $ -z\frac{dz}{dx} = z.$ $ z + x =c$ $\sqrt {x^2 - 4y}=c-x$ ${x^2 - 4y}=c^2+x^2 -2cx$ $4y = 2cx-c^2$ $y = \frac{cx}{2}-(\frac{c}{2})^2$ which is the equation of a line with gradient=c/2 and intercept on the X- axis also as c/2 This clearly shows $\frac{c}{2}=m$ and the intercept and gradient are same.

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    @GerryMyerson Gerry your second equation assumes cotA is constant, but cotA is a function of y'. Also the equation $xy'+y(1+m^2)=\sqrt{1+y'^2}$ when differentiated gives $ xm'+m+m+2mm'y+m^3=\frac{pmm'}{\sqrt{1+m^2}}$.Now putting $m'=0$ (considering a line) we get $m(2+m'^2)=0$ which shows either $m=0 $ or $2+m'^2=0$. $m=0 $ yields $y=c$ and the other has complex solution. Moreover the equation yields correct result for $y=p$ when we substitute $0$ for $m$. But if $m \rightarrow \propto $ it doesn't lead to the equation $x=p$. Instead yields $x+\propto=p$2012-07-08
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The answer to the second question is as follows . The lines in question are tangents to the circle with center at $(0,0)$

and radius=$p$. Now the length of the chord of a circle $x^2+y^2=p^2$ intercepted by straight line $y=mx+c$ is given by $2\sqrt\frac{p^2(1+m^2)-c^2}{1+m^2}$. Since for a tangent the length of the chord is $0$ we have $p^2(1+m^2)=c^2 $ This reduces the equation of the line to $ y=mx \pm p\sqrt{1+m^2}$ $(y-mx)^2=p^2(1+m^2) $$m^2(x^2-p^2)-2mxy+(y^2-p^2)=0 $

$m=\frac{xy\pm p\sqrt{x^2+y^2-p^2}}{(x^2-p^2)}$ Hence the equation of the line is given by the DE $\frac{dy}{dx}=\frac{xy\pm p\sqrt{x^2+y^2-p^2}}{(x^2-p^2)}$ Showing that $x=p$ when $m=\infty$ and $y=p$ when $m=0$ are solution to the DE

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The answer to your first question is as follows (I am using your solution only).

$2\frac{dy}{dx} - x = \sqrt {x^2 - 4y}$ Just solve for $m =\dfrac{dy}{dx}$ in your equation and you get the differential equation. Moreover note that when further differentiated we get $2\frac{d^2y}{dx^2} - 1 = (x-2y\frac{dy}{dx})/(\sqrt {x^2 - 4y})$ Which shows $2\frac{d^2y}{dx^2} - 1 = -1$ implying $\frac{d^2y}{dx^2}=0$ and that the DE gives the equation of curves with constant gradients which happens only in case of lines.

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    For $y=mx-m^2$ we have $2y'-x-\sqrt{x^2-4y}=2m-x-\sqrt{x^2-4mx+4m^2}=2m-x-|2m-x|,$ i.e. $y=mx-m^2$ does not solve the DE $2y'-x-\sqrt{x^2-4y}=0$.2012-06-30
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The first is correct. As for the second, you better not divide by $\sin A$ since the latter quantity may be zero.

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    There is certainly a mistake with the computation. However, that the DE has "many unwanted solutions" doesn't mean that the given function is NOT one of them!2012-06-29