# Find out the differential equation of the following families of curves.

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. –  Gerry Myerson Jun 27 '12 at 12:45
Please tell my mistake sir @GerryMyerson –  Bazinga Jun 27 '12 at 12:56
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. –  Gerry Myerson Jun 27 '12 at 13:08
@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! –  Mercy Jun 29 '12 at 19:34
@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 $y=mx-m^2$? –  Gerry Myerson Jun 29 '12 at 22:53

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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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 $\sqrt{x^2-4y}=\pm z$. –  Gerry Myerson Jul 1 '12 at 23:37
@GerryMyerson Both the equations are correct. One yields a gradient and X- axis intercept of c/2 and the other yields (-c/2) –  Barun Dasgupta Jul 2 '12 at 19:20
@GerryMyerson If delved further it would be seen that if we take $$2y'-x= \pm\sqrt{x^2 - 4y}$$ –  Barun Dasgupta Jul 7 '12 at 11:45
@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. –  Barun Dasgupta Jul 7 '12 at 14:40
@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 $x+\propto=p$ –  Barun Dasgupta Jul 8 '12 at 17:44

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 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. –  Gerry Myerson Jun 29 '12 at 22:56
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! –  Mercy Jun 29 '12 at 23:18

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 editing –  Barun Jun 28 '12 at 5:59
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 $2y'-x-\sqrt{x^2-4y}=0$. –  Mercy Jun 30 '12 at 11:30