Find the orthogonal trajectories of the family of curves given by $$x^2 + y^2 + 2Cy =1.$$
The ordinary differential equation for the family of curves is given by $y'=\frac{2xy}{x^2-y^2-1}$.Therefore, the differential equation for the orthogonal curves is given by $y'=\frac{1-x^2-y^2}{2xy}$.
This is an exact differential equation. So, solving by the standard method for exact differential equation gives $x-x^3/3+xy^2=C$. But this is not the correct answer according to the answers given at back of the book.
The answer given at the back of the book is $x^2 – y^2 – Cx +1 = 0$.
Can someone please find out at which step I am maing a mistake or provide a solution that leads to the correct answer?
Best Answer
For the given family of curves $$ x^2+y^2+2cy=1 $$ which can be written as $$ \frac{1-x^2-y^2}{2y} = c $$ differentiating both sides $$ \frac{dy}{dx}=\frac{2xy}{x^2-y^2-1} $$ and the family of curves orthogonal to this will be $$ -\frac{dx}{dy}=\frac{2xy}{x^2-y^2-1} $$ or, $$(x^2-y^2-1)dx+(2xy)dy=0 $$ this is not an exact DE, and hence will have to converted to one, $$ \frac{\partial{M}}{\partial{y}}=-2y; \frac{\partial{N}}{\partial{x}}=2y $$ $$ \frac{1}{N}\bigg(\frac{\partial{M}}{\partial{y}}-\frac{\partial{N}}{\partial{x}}\bigg)=-\frac{2}{x} $$ $$ I.F=e^{\int{-\frac{2}{x}}dx}=\frac{1}{x^2} $$ after multiply with the Integrating Factor the DE becomes exact $$ \bigg(1-\frac{y^2}{x^2}-\frac{1}{x^2}\bigg)dx+\frac{2y}{x}dx=0 $$ solving for this,the orthogonal family of curves is $$ x^2+y^2+1=cx $$