# Orthogonal Trajectories

#### Asawira Emaan

##### New member
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

Staff member
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
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

Staff member
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
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.