Question

When all the curves in a family$\mathbf{G}\left(x,y,c_1\right)\mathbf{=}\mathbf{0}$ intersect orthogonally all the curves in another family localid="1667974378968" $\mathbf{H}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}{\mathbf{c}}_{\mathbf{1}}\mathbf{)}\mathbf{=}\mathbf{0}$, the families are said to be orthogonal trajectories of each other. See Figure 3.R.5. If localid="1667974383247" $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$ is the differential equation of one family, then the differential equation for the orthogonal trajectories of this family is localid="1667974387852" $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$.Find the differential equation of the given family by computing localid="1667974392147" $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}$ and eliminating localid="1667974397114" ${\mathbf{c}}_{\mathbf{1}}$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.

$\mathit{y}\mathbf{}\mathbf{=}\mathbf{}{\mathit{c}}_{\mathbf{1}}\mathit{x}$

Short answer

The orthogonal trajectories of the given family is $y=c_{1}x$ and$y^2+x^2=c$

Step-by-step solution

Compute dxdy for the two families

We have the family

$y=c_{1}x$

Which its curves intersect orthogonally all the curves of another family.

First, we have to find the value of constant $c_1$from equation (1) as

$c_1=\frac{y}{x}$

Second, we have to derivative the family shown in equation (1) with respect to $x$, then we have

$\frac{dy}{dx}=c_1$

Then, from (2) and (3), we can have

$\frac{dy}{dx}=\frac{y}{x}$is the differential equation of the given family $y-c_{1}x$

Third, we can obtain the differential equation of the orthogonal trajectories of the given family as

$$\begin{gathered} \frac{dy}{dx}=-\frac{1}{{\displaystyle \frac{y}{x}}} \\ =-\frac{x}{y} \end{gathered}$$

Find the condition

Finally, since this differential equation shown in (5) is separable, then we can solve it as

$$\begin{gathered} \int ydy=-\int xdx \\ \frac{1}{2}{y}^2=-\frac{1}{2}{x}^2+c_2 \\ {y}^2+x^2=c \end{gathered}$$

is the orthogonal trajectories of the given family.

Then from (1) and (6), we can draw the given family and the orthogonal trajectories family as

Therefore, the orthogonal trajectories of the given family is $y=c_{1}x$and $y^2+x^2=c$