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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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    $y'=m$ gives all lines of slope $m$, not just the ones with slope equal to intercept.2012-06-27
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    Please tell my mistake sir @GerryMyerson2012-06-27
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    I thought I did tell you your mistake. You wanted to get lines of the form $y=mx-m^2$. But $y'=m$ gives lines of the form $y=mx+c$, whether $c$ is $-m^2$ or not.2012-06-27
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    @GerryMyerson His purpose is to find a differential equation satisfied by the line $y=mx-m^2$ NOT to find the line from the differential equation!2012-06-29
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    @Mercy, in that case, $y''=0$ would be a perfectly good answer, or $y'''=0$, or.... Why would OP specify "slope and $x$-intercept equal in magnitude" unless OP wanted a differential equation satisfied *only* by lines of the form $y=mx-m^2$?2012-06-29
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    @GerryMyerson The problem is to find a DE satisfied by a function, NOT to find a specific solution of a DE which requires a condition anyway. A problem whose solution is $y=mx-m^2$ is NOT just a DE because one needs to specify e.g. $y(0)$, etc.2012-06-29
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    @Mercy, you've lost me. Why isn't $y'''=0$ as good an answer as $y'=m$ to OP's first question? or do you think it is just as good an answer?2012-06-29
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    @GerryMyerson Where did I say that $y=mx-m^2$ does not solve $y''=0$? Please show?2012-06-29
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    @Mercy, then I take it you think $y''=0$ is a perfectly good answer to the question, as is $y'''=0$ and for that matter $(y-mx+m^2)y'=0$, since the solutions to those equations all include the lines OP wants. You might be right, but if you are then it's a ridiculous question. It just has too many correct answers, most of which completely ignore the bit about "slope and $x$-intercept equal in magnitude."2012-06-30
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    @GerryMyerson Here's OP's question: Are my answers correct? Here's what you wrote "I thought I did tell you your mistake. You wanted to get lines of the form $y=mx−m^2$. But $y'=m$ gives lines of the form $y=mx+c$, whether $c$ is $−m^2$ or not", so my question is: Was his purpose to get the lines of form $y=mx-m^2$? No!, and does $y=mx-m^2$ solve the DE $y'=m$? YES! I would be very glad if you could provide a 1st order linear DE satisfied only by $y=mx-m^2$.2012-06-30
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    @Mercy, I think that Barun has already provided a 1st order DE satisfied only by $y=mx-m^2$ in an answer posted two days ago. It is not linear, but OP didn't ask for that (and I don't know if it's possible).2012-06-30
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    @GerryMyerson Have you checked that $y=mx-m^2$ is a solution of the DE provided by Barun? For $y=mx-m^2$ one actually has:$$2y'-x-\sqrt{x^2-4y}=2m-x-\sqrt{x^2-4mx+4m^2}=2m-x-|2m-x|$$ which is obviously not zero!2012-06-30
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    @Mercy, it is if you take $\sqrt{x^2-4mx+4m^2}=2m-x$.2012-06-30
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    @GerryMyerson Try to be honest! $\sqrt{a^2}=a$ only if $a \ge 0$! and $y=mx-m^2$ is defined for every $x \in \mathbb{R}$, therefore your identity makes no sense for $x>2m$.2012-06-30
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    @Mercy, I would have been happier if Barun had not taken the square root, and had instead simply written $y=xy'-(y')^2$.2012-06-30
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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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    You didn't say how you got to the second formula - the one in the edit...2017-12-16
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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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    Mercy will object that $z^2=x^2-4y$ does not imply $\sqrt{x^2-4y}=z$ but instead $\sqrt{x^2-4y}=\pm z$.2012-07-01
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    @GerryMyerson Both the equations are correct. One yields a gradient and X- axis intercept of c/2 and the other yields (-c/2)2012-07-02
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    @GerryMyerson If delved further it would be seen that if we take $$ 2y'-x= \pm\sqrt{x^2 - 4y}$$2012-07-07
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    @GerryMyerson The solution (DE) $2\frac{dy}{dx} - x = \sqrt {x^2 - 4y}$ to the first question actually leads to a quadratic equation of the type $ y = ax^2+bx+(\frac{b}{2a-1})^2(a-1) $ which makes its gradients and x intercepts equal at certain points. For special cases putting a =0, a=1,b=0 and a=1/2 we get four type of curves $y=bx-b^2$, $y= x^2+bx$, $y=cx^2 $, $y=0.5x^2+c$. This shows that appart from straight lines of the type $ y=bx-b^2$ we have certain other curves also that can be derived from this DE and which satisfy the general condition of x-intercept and gradient equality.2012-07-07
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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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    Peter , Thanks for your editing2012-06-28
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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's a bigger problem than that. For one thing, dividing by $\sin A$ and getting $p$ *times* $\sin A$ on the right. For another, the same objection as for the first answer, that the differential equation has many unwanted solutions.2012-06-29
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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