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


*

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

*Straight lines at a fixed distance $p$ from the origin.
My Approach :


*

*a straight line is defined by $y = mx + c$, $x$-intercept $= -c/m$ 
but $-c/m = m$, so $c = -m^2$. 
$$y = mx - m^2\; , \; dy/dx = m\; , \;  y' = m$$

*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?
 A: 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.  
A: 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.
A: 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
A: 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.
A: The first is correct. As for the second, you better not divide by $\sin A$ since the latter quantity may be zero.
