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.

\(y = {c_1}{e^x}\)

Short answer

\(\frac{{{y^2}}}{2} = - x + {c_2}\)

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\(y = {c_1}{e^x}\,\,\, - - - (1)\)

Differentiate equation (1) w.r.t\(x\).

\(\begin{array}{l}\frac{{dy}}{{dx}} = {c_1}{e^x}\\\frac{{dy}}{{dx}} = y\left( {{\rm{ since }}y = {c_1}{e^x}} \right)\end{array}\)

So, differential equation of the given family is\(\frac{{dy}}{{dx}} = y\).

Differential equation of orthogonal trajectory

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

\(\begin{array}{l}\frac{{dy}}{{dx}} = - \frac{1}{{f(x,y)}}\\\frac{{dy}}{{dx}} = \frac{{ - 1}}{y}\end{array}\)

Solve the above equation.

\(\begin{array}{l}\int y dy = \int - dx\\\frac{{{y^2}}}{2} = - x + {c_2}\end{array}\)

Orthogonal trajectories of the given family is\(\frac{{{y^2}}}{2} = - x + {c_2}\).