What are the orthogonal trajectories of e^{x}(xcosy - ysiny) = c?

[Asawira Emaan]

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]

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:

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

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

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

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

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

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

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

And so, the orthogonal trajectories will satisfy:

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

Let's express this in differential form:

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

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

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

Thus, we may write:

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

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

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

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

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

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

Or:

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

Hence, the solution is given implicitly by:

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

Back substitute for \(u\):

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

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

 

[Alone]

We are given the family of curves:

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

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

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

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

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

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

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

And so, the orthogonal trajectories will satisfy:

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

Let's express this in differential form:

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

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

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

Thus, we may write:

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

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

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

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

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

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

Or:

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

Hence, the solution is given implicitly by:

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

Back substitute for \(u\):

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

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

How did you get the LHS of:
$$e^u\left(u\cos(y)-y\sin(y)\right)=e^u(u-1)\cos(y)+e^u(\cos(y)-y\sin(y))+g'(y)$$

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]

How did you get the LHS of:
$$e^u\left(u\cos(y)-y\sin(y)\right)=e^u(u-1)\cos(y)+e^u(\cos(y)-y\sin(y))+g'(y)$$

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:

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

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

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

and

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

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

 

[Alone]

Suppose we have an ODE in differential form and that it is exact:

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

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

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

and

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

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.