Question
Which one of the following differential equations represents the family of straight lines which are at unit distance from the origin?
A.
$${\left( {y - x\frac{{dy}}{{dx}}} \right)^2} = 1 - {\left( {\frac{{dy}}{{dx}}} \right)^2}$$
B.
$${\left( {y + x\frac{{dy}}{{dx}}} \right)^2} = 1 + {\left( {\frac{{dy}}{{dx}}} \right)^2}$$
C.
$${\left( {y - x\frac{{dy}}{{dx}}} \right)^2} = 1 + {\left( {\frac{{dy}}{{dx}}} \right)^2}$$
D.
$${\left( {y + x\frac{{dy}}{{dx}}} \right)^2} = 1 - {\left( {\frac{{dy}}{{dx}}} \right)^2}$$
Answer :
$${\left( {y - x\frac{{dy}}{{dx}}} \right)^2} = 1 + {\left( {\frac{{dy}}{{dx}}} \right)^2}$$
Solution :
$$y = mx + c$$ (Equation of straight line)
$$\frac{{dy}}{{dx}} = m$$ and $$mx - y + c = 0$$ is at unit distance from origin.
$$\eqalign{
& \therefore \,\frac{{\left| {m\left( 0 \right) - \left( 0 \right) + c} \right|}}{{\sqrt {{m^2} + {{\left( { - 1} \right)}^2}} }} = 1\, \Rightarrow c = \sqrt {1 + {m^2}} \cr
& {\text{Now}}\,{\text{:}}\,{\left[ {y - x\frac{{dy}}{{dx}}} \right]^2} = {\left[ {mx + c - xm} \right]^2} = {c^2} = 1 + {m^2} \cr
& {\text{also,}}\,{\left[ {y + x\frac{{dy}}{{dx}}} \right]^2} = {\left[ {mx + c + xm} \right]^2} = {\left[ {2mx + \sqrt {1 + {m^2}} } \right]^2} \cr
& {\text{also,}}\,1 - {\left( {\frac{{dy}}{{dx}}} \right)^2} = 1 - {m^2}{\text{ and }}1 + {\left( {\frac{{dy}}{{dx}}} \right)^2} = 1 + {m^2} \cr
& \Rightarrow {\left[ {y - x\frac{{dy}}{{dx}}} \right]^2} = 1 + {\left( {\frac{{dy}}{{dx}}} \right)^2} \cr} $$