Question

Solution of the differential equation $\left( {\cos x} \right)dy = y\left( {\sin x - y} \right)dx,0 < x < \dfrac{\pi }{2}$ is:
A) $y\sec x = \tan x + c$
B) $y\tan x = \sec x + c$
C) $\tan x = \left( {\sec x + c} \right)y$
D) $\sec x = \left( {\tan x + c} \right)y$

Answer 1

Hint: Write the equation in the form of $\dfrac{{dy}}{{dx}} + f\left( x \right) \times y = g\left( x \right)$ and then find the integrating factor
$IF = {e^{\int {f\left( x \right)dx} }}$ and we know the general equation will become
$\left( {IF} \right)y = \int {\left( {IF} \right)g\left( x \right)dx + c} $

Complete step-by-step answer:
Here we are given a differential equation
$\left( {\cos x} \right)dy = y\left( {\sin x - y} \right)dx$
So, we arrange it in the form of $\dfrac{{dy}}{{dx}} + f\left( x \right) \times y = g\left( x \right)$.
So, we get,
$\dfrac{{dy}}{{dx}} = \dfrac{{y\sin x - {y^2}}}{{\cos x}}$
So, this cannot be in the form of $\dfrac{{dy}}{{dx}} + f\left( x \right) \times y = g\left( x \right)$. If we divide by ${y^2}$ both side,
$\dfrac{1}{{{y^2}}}\dfrac{{dy}}{{dx}} = \dfrac{1}{y}\dfrac{{\sin x}}{{\cos x}} - \dfrac{1}{{\cos x}}$
$\dfrac{1}{{{y^2}}}\dfrac{{dy}}{{dx}} - \dfrac{1}{y} \times \tan x = - \sec x$ (1)
Now we can assume that $ - \dfrac{1}{y} = t$
Upon differentiation with respect to $x$, we get
$\dfrac{1}{{{y^2}}}\dfrac{{dy}}{{dx}} = \dfrac{{dt}}{{dx}}$
Now we can replace, $\dfrac{1}{{{y^2}}}\dfrac{{dy}}{{dx}} = \dfrac{{dt}}{{dx}}$ and $ - \dfrac{1}{y} = t$ in equation (1)
So, we get
$\dfrac{{dt}}{{dx}} + t\tan x = - \sec x$
So we get in the form of $\dfrac{{dy}}{{dx}} + f\left( x \right) \times y = g\left( x \right)$.
So, now we need to find the integrating factor.
$
  IF = {e^{\int {f\left( x \right)dx} }} \\
  {\text{Here, }}f\left( x \right) = \tan x \\
  IF = {e^{\int {\tan xdx} }} = {e^{{{\log }_e}\left( {\sec x} \right)}} = \sec x \\
  {\text{so, here we got }}IF = \sec x \\
 $
Now the general equation will become
$\left( {IF} \right)y = \int {\left( {IF} \right)g\left( x \right)dx + c} $
Here replace $y$ by $t$
$
  \left( {\sec x} \right)t = \int { - \sec x \times \sec xdx} + c \\
  \left( {\sec x} \right)t = - \int {{{\sec }^2}xdx} + c \\
    \\
 $
We know that $\int {{{\sec }^2}xdx} = \tan x$
So,
$t\left( {\sec x} \right) = - \tan x + c$
And we know $ - \dfrac{1}{y} = t$
$
  \dfrac{{\sec x}}{y} = \tan x + c \\
  \sec x = \left( {\tan x + c} \right)y \\
 $

So, option D is correct.

Note: Here we can use alternative method, we are given $\left( {\cos x} \right)dy = y\left( {\sin x - y} \right)dx$
$ \Rightarrow \cos xdy - y\sin xdx = - {y^2}dx$
If we differentiate $d\left( {y\cos x} \right) = \cos xdy - y\sin xdx$
So, $d\left( {y\cos x} \right) = - {y^2}dx$
Now multiplying and dividing by ${\cos ^2}x$
$
   \Rightarrow d\left( {y\cos x} \right) = \dfrac{{ - {y^2}{{\cos }^2}x}}{{{{\cos }^2}x}}dx \\
  \dfrac{{d\left( {y\cos x} \right)}}{{{y^2}{{\cos }^2}x}} = - \dfrac{{dx}}{{{{\cos }^2}x}} \\
 $
Integrating both sides,
$
  \int {\dfrac{1}{{{{\left( {y\cos x} \right)}^2}}}d\left( {y\cos x} \right)} = - \int {{{\sec }^2}xdx} \\
   - \dfrac{1}{{y\cos x}} = - \tan x + c \\
  \sec x = \left( {\tan x + c} \right)y \\
 $