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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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.

Answer 5

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