Question

Equation of a family of curves is $\sin y = k{e^{{x^2}}}$ , what is the differential equation of the family with its orthogonal trajectory?
A.$\dfrac{{dy}}{{dx}} = - \dfrac{{\tan y}}{{2x}}$
B.$\dfrac{{dy}}{{dx}} = - \dfrac{{\sin y}}{{2x}}$
C.$\dfrac{{dy}}{{dx}} = - \dfrac{{\cos y}}{{2x}}$
D.$\dfrac{{dy}}{{dx}} = - \dfrac{{\cot y}}{{2x}}$

Answer 1

Hint: In this question we have been given $\sin y = k{e^{{x^2}}}$ . First of all we will differentiate both the sides of the equation with respect to $x$ .
We know that the derivative of sine is cosecant i.e.
$\dfrac{{dy}}{{dx}}\sin y = \cos y$ .
 After that we will substitute $ - \dfrac{{dx}}{{dy}}$ in place for $\dfrac{{dy}}{{dx}}$ in the differential equation. This will give us the differential equation of the orthogonal trajectories.

Complete answer:In this question, we have
$\sin y = k{e^{{x^2}}}$.
We will differentiate both the sides of the equation with respect to $x$ .
 We know the value of $\dfrac{{dy}}{{dx}}\sin y$ is
 $\cos y$ .
And the derivative of $k{e^{{x^2}}}$ is
$2xk{e^{{x^2}}}$ .
By putting the values in the equation, we have:
$\cos y\dfrac{{dy}}{{dx}} = 2xk{e^{{x^2}}}$ .
We can write $\sin y$ in place of $k{e^{{x^2}}}$ , it is given in the question.
So we have
$\cos y\dfrac{{dy}}{{dx}} = 2x\sin y$ .
Now we will substitute $ - \dfrac{{dx}}{{dy}}$ with $\dfrac{{dy}}{{dx}}$ in the above equation, it gives:
$ - \cos y\dfrac{{dx}}{{dy}} = 2x\sin y$
By taking $ - \cos y$ to the right hand side of the equation, we have:
$\dfrac{{dx}}{{dy}} = - \dfrac{{2x\sin y}}{{\cos y}}$
We will reciprocate the values in both sides of the equation i.e. $\dfrac{{dx}}{{dy}}$ can be written as $\dfrac{{dy}}{{dx}}$
And,
$ - \dfrac{{2x\sin y}}{{\cos y}} = - \dfrac{{\cos y}}{{2x\sin y}}$
So we have
 $\dfrac{{dy}}{{dx}} = - \dfrac{{\cos y}}{{2x\sin y}}$
Now we know that
 $\dfrac{{\cos y}}{{\sin y}} = \cot y$ , by putting this value in the equation it gives:
$\dfrac{{dy}}{{dx}} = - \dfrac{{\cot y}}{{2x}}$
Hence the correct option is (d) $\dfrac{{dy}}{{dx}} = - \dfrac{{\cot y}}{{2x}}$

Note:
We should note that in the above solution $k{e^{{x^2}}}$ , we have $k$ as the constant. So initially we have to differentiate ${e^{{x^2}}}$ .
We will apply the chain rule here i.e.
 $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$
Here we have assumed
$y = {e^u}$ and
 $u = {x^2}$ .
So by applying the formula we can write:
 $\dfrac{d}{{du}}{e^u} \times \dfrac{d}{{dx}}({x^2})$
We will again apply the power rule which says that
 $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ , where n is the exponential power.
Here we have
 $n = 2$
So we can write
 $2{x^{2 - 1}} = 2x$
By putting the value, in the equation we have:
${e^u} \times 2x$
And, from the above we have assumed
$u = {x^2}$ .
So we can write it as
${e^2} \times 2x = 2x{e^{{x^2}}}$
By putting the constant back in the equation, we have
$2xk{e^{{x^2}}}$ .