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?

  • 2
    $\begingroup$ Wouldn't it be easier to complete the square first? $\endgroup$
    – abiessu
    Jun 15, 2014 at 14:26
  • $\begingroup$ Your equation for the family of curves is not correct. Can you show the steps of your solution? $\endgroup$
    – Artem
    Jun 15, 2014 at 17:23

2 Answers 2


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 $$ x^2+y^2+1=cx $$

  • $\begingroup$ Could you explain how you get 3rd equation from the 2nd? What did you differentiate? $\endgroup$ May 3, 2019 at 5:34
  • $\begingroup$ make the equation such that y^2 = something, then you can differentiate both sides to get 2y * dy/dx = d(something)/dx ; then just rearrange the terms. $\endgroup$ Jun 29, 2019 at 16:09

The answer at the end of your book does not seem correct. The correct one is $y^2 + x^2 - Cx + 1 = 0$.

The equation is solved most easily with the substitution (all derivatives in Lagrange notation are with respect to $x$):

$$ (y^2 + x^2 + 2Cy)' = (1)' \quad \Rightarrow \quad 2yy' + 2x + 2Cy' = 0 $$

Express $2C$ with the help of the initial equation:

$$y^2 + x^2 + 2Cy = 1 \quad \Rightarrow \quad 2C = \frac{1 - x^2 - y^2}{y}$$

Substitute this into the derivative equation:

$$2yy' + 2x + \frac{1 - x^2 - y^2}{y}y' = 0 \quad \Rightarrow \quad 2y^2y' + 2xy + y' -x^2y' - y^2y' = 0, y \neq 0$$

where we removed the $y$ in the denominator, by restricting the sought function $y$ to be nonzero. Now grouping the terms:

$$2xy + (y^2 - x^2 + 1)y' = 0 \quad \Rightarrow \quad y' = \frac{-2xy}{y^2 - x^2 + 1}$$

The family of orthogonal curves would have their direction field as a reciprocal of the direction field we found above^. Thus it can be expressed as $y_0$:

$$y_0' = \frac{y_0^2 - x^2 + 1}{2xy_0} \quad \Rightarrow \quad 2xy_0y_0' = y_0^2 - x^2 + 1$$

It is easy to notice now, that if we take $v = y_0^2$. then this expression is equivalent to:

$$xv' = v - x^2 + 1 \quad \Rightarrow \quad v' - \frac{1}{x}v = \frac{1}{x} - x$$

But this is the first order differential equation for which we have an existence uniqueness theorem, and a formula, where $P(x) = -\frac{1}{x}$, and $Q(x) = \frac{1}{x} - x$, and $A(x) = \int_1^x P(t)dt = -\log(x)$. The point $x = 1$ is selected as an initial value point for simplicity, since the logarithm which appears in the integration becomes 0 there. Moreover, the fact that the term $1/x$ is integrated from 1 means that the integral is not defined at $x = 0$, so if the initial value is taken positive, the solution will be only defined for $x > 0$. This allows to remove the absolute value from the logarithm when after integration. The solution is then given by:

$$v = v(1)e^{-A(x)} + e^{-A(x)} \int_{1}^x e^{A(t)}Q(t)dt$$


$$v = v(1)e^{\log(x)} + e^{\log(x)}\int_1^x (\frac{1}{t^2} - 1)dt = v(1)x + x(-\frac{1}{x} + 1 - x + 1) = v(1)x - 1 + 2x - x^2$$

Now, $v(1)$ is the initial condition of the equation, which we assume to be some constant $C$. We group it with other similar terms:

$$v = (v(1) + 2)x - x^2 - 1 = -x^2 + Cx - 1$$

However, we remember that we substituted $v = y_0^2$. Thus:

$$y_0^2 = -x^2 + Cx - 1 \quad \Rightarrow \quad y_0^2 + x^2 - Cx + 1 = 0$$

which is the required answer.

PS. I do not know if this linear ODE solution formula is common knowledge, but it directly follows from the proof of uniqueness/existence theorem and sure is defined as such in Apostol calculus 1 page 310.


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