Question

Question: When all the curves in a family \(G\left( {x,y,{c_1}} \right) = 0\)intersect orthogonally all the curves in another family\(H\left( {x,y,{c_2}} \right) = 0\), the families are said to be orthogonal trajectories of each other. See Figure 3.R.5. If \(dy/dx = f(x,y)\)is the differential equation of one family, then the differential equation for the orthogonal trajectories of this family is\(dy/dx = - 1/f(x,y)\). In Problems\(\;{\bf{15}} - {\bf{18}}\)find the differential equation of the given family by computing\(dy/dx\)and eliminating \({c_1}\)from this equation. Then find the orthogonal trajectories of the family. Use a graphing utility to graph both families on the same set of coordinate axes.

\(y = \frac{1}{{x + {c_1}}}\)

Short answer

\({y^3} = 3x + {c_2}\)

Step-by-step solution

Definition

Orthogonal trajectory family of curves that intersect another family of curves at right angles.

Differential equation of given family

Consider\(y = \frac{1}{{x + {c_1}}}\,\,\,\,\, - - - - - (1)\)

Differentiate both sides with respect to\(x\),

\(\begin{array}{l}\frac{{dy}}{{dx}} = - \frac{1}{{{{\left( {x + {c_1}} \right)}^2}}}\\\frac{{dy}}{{dx}} = - {y^2}\left( {{\rm{ Use }}y = \frac{1}{{x + {c_1}}}} \right)\end{array}\)

Compare it with\(dy/dx = f(x,y)\), to get\(f(x,y) = - {y^2}\).

Differential equation of orthogonal trajectory

Differential equation for the orthogonal trajectories of the given family is:

In this case, \(f(x,y) = - {y^2}\)

\(\begin{array}{c}\frac{{dy}}{{dx}} = \frac{{ - 1}}{{\left( { - {y^2}} \right)}}\\\frac{{dy}}{{dx}} = \frac{1}{{{y^2}}}\\{y^2}dy = dx\\\frac{1}{3}{y^3} = x + {c_0}\;\;\;\left( {{\rm{ Here }}{c_2} = 3{c_0}} \right)\end{array}\)

Hence, orthogonal trajectories of the family \(y = \frac{1}{{x + {c_1}}}\) is\({y^3} = 3x + {c_2}\).