OrthogonalTrayectories::usage="When all the curves in a family G(x, y,
c
1
​) = 0 intersect orthogonally with all the curves of the other family H(x, y,
c
2
​) = 0,​ the families are said to be orthogonal trajectories of each other. If
y
x
= f(x, y) it's the differential equation of a family, then the differential equation for the orthogonal trajectories of this family will be
y
x
= -
1
f(x,y)
. Find the orthogonal trajectories for the family y =
1
(x+
c
1
)
​ References: Zill, D.G., Cullen, M.R. Differential Equations with ​​ Boundary-Value Problems. Ed. Brooks/Cole: p.80, 1977.";​​
Development:
xy+
c
1
y=1
We derive implicity with respect to x:
x
y
x
+y+
c
1
y
x
=0;​​
y
x
=-
y
x+
c
1
;
From the original equation we clear
c
1
and we substitute it in the differential equation:
c
1
=
1
y
-x;​​
y
x
=-
y
x+
1
y
-x
;
y
x
=-
2
y
;​​
We look for the orthogonal trajectories, then:
y
x
=
1
2
y
;
In[]:=
DSolvey'[x]==1
2
(y[x])
,y[x],x
Out[]=
y[x]-
1/3
(-3)
1/3
(x+

1
)
,y[x]
1/3
3
1/3
(x+

1
)
,y[x]
2/3
(-1)
1/3
3
1/3
(x+

1
)

We take one of the solutions and raise it to the cube:
3
y
=3x+
c
2
In[]:=
ContourPloty==
1
x-5
,
3
y
==3x-5,y==
1
x-4
,
3
y
==3x-4,y==
1
x-3
,
3
y
==3x-3,y==
1
x-2
,
3
y
==3x-2,y==
1
x-1
,
3
y
==3x-1,y==
1
x
,
3
y
==3x,y==
1
x+1
,
3
y
==3x+1,y==
1
x+2
,
3
y
==3x+2,y==
1
x+3
,
3
y
==3x+3,y==
1
x+4
,
3
y
==3x+4,y==
1
x+5
,
3
y
==3x+5,{x,-5,5},{y,-5,5},AxesTrue,AxesOrigin{0,0},BoxRatiosAutomatic,PlotRangeAll,PerformanceGoal"Quality"
Out[]=
In[]:=
NSolvey==
1
x-2
,
3
y
==3x-2,{x,y}
Out[]=
{{x2.56043,y1.78436},{x1.77514+0.601106,y-0.545927-1.45938},{x1.77514-0.601106,y-0.545927+1.45938},{x0.555967,y-0.692505}}
In[]:=
δ=ResourceFunction["TangentLine"]y==
1
x-2
,{x,2.560425660450319},y,"Plot";​​ζ=ResourceFunction["TangentLine"][
3
y
==3x-2,{x,2.560425660450319},y,"Plot"];​​Show[δ,ζ]​​​​
Out[]=
This graph shows at the point {2.56043, 1.78436} para las curvas con
c
1
=
c
2
=-2
clearly because they are called orthogonal trajectories. Their tangent lines at that point are orthogonal to each other.
Here are the equations of the tangent lines in each case:
In[]:=
ResourceFunction["TangentLine"]y==
1
x-2
,{x,2.560425660450319},y,"SlopeInterceptEquation"
Out[]=
y9.93658-3.18393x
In[]:=
ResourceFunction["TangentLine"][
3
y
==3x-2,{x,2.560425660450319},y,"SlopeInterceptEquation"]
Out[]=
y0.980187+0.314077x