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Orthogonal Trajectories

Asawira Emaan

New member
Nov 14, 2018
12
Asalamoalaikum, help me with this. I can solve it but it goes very lengthy.
Determine the equations of the orthogonal trajectories of the following family of curve;
e^{x}(xcosy - ysiny) = c
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Asalamoalaikum, help me with this. I can solve it but it goes very lengthy.
Determine the equations of the orthogonal trajectories of the following family of curve;
e^{x}(xcosy - ysiny) = c
We are given the family of curves:

\(\displaystyle e^x\left(x\cos(y)-y\sin(y)\right)=c\)

I would begin by implicitly differentiating w.r.t \(x\):

\(\displaystyle e^x\left(x\cos(y)-y\sin(y)\right)+e^x\left(\cos(y)-x\sin(y)y'-y'\sin(y)-y\cos(y)y'\right)=0\)

Since \(e^x\ne0\) divide through by that, and solve for \(y'\) to obtain:

\(\displaystyle y'=\frac{(x+1)\cos(y)-y\sin(y)}{(x+1)\sin(y)+y\cos(y)}\)

Let \(u=x+1\implies du=dx\)

\(\displaystyle y'=\frac{u\cos(y)-y\sin(y)}{u\sin(y)+y\cos(y)}\)

And so, the orthogonal trajectories will satisfy:

\(\displaystyle y'=\frac{u\sin(y)+y\cos(y)}{y\sin(y)-u\cos(y)}\)

Let's express this in differential form:

\(\displaystyle \left(u\sin(y)+y\cos(y)\right)du+\left(u\cos(y)-y\sin(y)\right)dy=0\)

This equation is not exact, but we easily derive the integrating factor:

\(\displaystyle \mu(u)=e^u\)

Thus, we may write:

\(\displaystyle e^u\left(u\sin(y)+y\cos(y)\right)du+e^u\left(u\cos(y)-y\sin(y)\right)dy=0\)

We indeed find that this equation is exact, and so we must have:

\(\displaystyle F(x,y)=\int e^u\left(u\sin(y)+y\cos(y)\right)du+g(y)\)

\(\displaystyle F(x,y)=e^u(u-1)\sin(y)+e^uy\cos(y)+g(y)\)

Taking the partial derivative w.r.t \(y\), we obtain:

\(\displaystyle e^u\left(u\cos(y)-y\sin(y)\right)=e^u(u-1)\cos(y)+e^u(\cos(y)-y\sin(y))+g'(y)\)

Or:

\(\displaystyle g'(y)=0\implies g(y)=C\)

Hence, the solution is given implicitly by:

\(\displaystyle e^u((u-1)\sin(y)+y\cos(y))=C\)

Back substitute for \(u\):

\(\displaystyle e^{x+1}(x\sin(y)+y\cos(y))=C\)

This is the family of orthogonal trajectories to the given family of curves. :)
 

Alan

Member
Jul 21, 2012
60
We are given the family of curves:

\(\displaystyle e^x\left(x\cos(y)-y\sin(y)\right)=c\)

I would begin by implicitly differentiating w.r.t \(x\):

\(\displaystyle e^x\left(x\cos(y)-y\sin(y)\right)+e^x\left(\cos(y)-x\sin(y)y'-y'\sin(y)-y\cos(y)y'\right)=0\)

Since \(e^x\ne0\) divide through by that, and solve for \(y'\) to obtain:

\(\displaystyle y'=\frac{(x+1)\cos(y)-y\sin(y)}{(x+1)\sin(y)+y\cos(y)}\)

Let \(u=x+1\implies du=dx\)

\(\displaystyle y'=\frac{u\cos(y)-y\sin(y)}{u\sin(y)+y\cos(y)}\)

And so, the orthogonal trajectories will satisfy:

\(\displaystyle y'=\frac{u\sin(y)+y\cos(y)}{y\sin(y)-u\cos(y)}\)

Let's express this in differential form:

\(\displaystyle \left(u\sin(y)+y\cos(y)\right)du+\left(u\cos(y)-y\sin(y)\right)dy=0\)

This equation is not exact, but we easily derive the integrating factor:

\(\displaystyle \mu(u)=e^u\)

Thus, we may write:

\(\displaystyle e^u\left(u\sin(y)+y\cos(y)\right)du+e^u\left(u\cos(y)-y\sin(y)\right)dy=0\)

We indeed find that this equation is exact, and so we must have:

\(\displaystyle F(x,y)=\int e^u\left(u\sin(y)+y\cos(y)\right)du+g(y)\)

\(\displaystyle F(x,y)=e^u(u-1)\sin(y)+e^uy\cos(y)+g(y)\)

Taking the partial derivative w.r.t \(y\), we obtain:

\(\displaystyle e^u\left(u\cos(y)-y\sin(y)\right)=e^u(u-1)\cos(y)+e^u(\cos(y)-y\sin(y))+g'(y)\)

Or:

\(\displaystyle g'(y)=0\implies g(y)=C\)

Hence, the solution is given implicitly by:

\(\displaystyle e^u((u-1)\sin(y)+y\cos(y))=C\)

Back substitute for \(u\):

\(\displaystyle e^{x+1}(x\sin(y)+y\cos(y))=C\)

This is the family of orthogonal trajectories to the given family of curves. :)
How did you get the LHS of:
\(\displaystyle e^u\left(u\cos(y)-y\sin(y)\right)=e^u(u-1)\cos(y)+e^u(\cos(y)-y\sin(y))+g'(y)\)

The RHS of the above is $F'_y$, that's clear to me but I don't understand how did you get the LHS?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
How did you get the LHS of:
\(\displaystyle e^u\left(u\cos(y)-y\sin(y)\right)=e^u(u-1)\cos(y)+e^u(\cos(y)-y\sin(y))+g'(y)\)

The RHS of the above is $F'_y$, that's clear to me but I don't understand how did you get the LHS?
Suppose we have an ODE in differential form and that it is exact:

\(\displaystyle M(x,y)\,dx+N(x,y)\,dy=0\)

Now, because the ODE is exact, this implies that there is a function \(F(x,y)\) such that:

\(\displaystyle \pd{F}{x}=M(x,y)\)

and

\(\displaystyle \pd{F}{y}=N(x,y)\)

And it is this implication of an exact equation that gave me the LHS of that equation. :)
 

Alan

Member
Jul 21, 2012
60
Suppose we have an ODE in differential form and that it is exact:

\(\displaystyle M(x,y)\,dx+N(x,y)\,dy=0\)

Now, because the ODE is exact, this implies that there is a function \(F(x,y)\) such that:

\(\displaystyle \pd{F}{x}=M(x,y)\)

and

\(\displaystyle \pd{F}{y}=N(x,y)\)

And it is this implication of an exact equation that gave me the LHS of that equation. :)
I see, I forgot about it.
Thanks for reminding me this.