Asawira Emaan said: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
How did you get the LHS of:MarkFL said: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 said: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?
I see, I forgot about it.MarkFL said: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. :)
The equation for the orthogonal trajectories is given by y = -cot(x) + c/e^{x}.
To find the orthogonal trajectories, we first rewrite the given equation in the form y = f(x). Then, we find the derivative of f(x) and take its negative reciprocal. This gives us the slope of the orthogonal trajectories. Finally, we integrate the slope to find the equation of the orthogonal trajectories.
The orthogonal trajectories represent a family of curves that intersect the given curve at right angles. They are perpendicular to the given curve at every point of intersection.
No, the orthogonal trajectories can only intersect the given curve once at each point. This is because the slope of the orthogonal trajectories changes sign at each point of intersection, resulting in only one point of intersection.
The value of c determines the specific curve within the family of orthogonal trajectories. Different values of c will result in different curves that still intersect the given curve at right angles.