Question

Solve: $\sin 2x\left( {\dfrac{{dy}}{{dx}} - \sqrt {\tan x} } \right) - y = 0$

Answer 1

Hint: In order to solve this equation, we will divide the whole equation by $\sin 2x$ and make it much simpler. By doing this, we will be eliminating the $\sin 2x$ in the starting of the equation which will result in a much simpler solution.

Complete step-by-step answer:
Diving the whole equation $\sin 2x\left( {\dfrac{{dy}}{{dx}} - \sqrt {\tan x} } \right) - y = 0$ by $\sin 2x$, we will get-
$
  \dfrac{{\sin 2x\left( {\dfrac{{dy}}{{dx}} - \sqrt {\tan x} } \right) - y}}{{\sin 2x}} = 0 \\
    \\
  \sin 2x\dfrac{{\left( {\dfrac{{dy}}{{dx}} - \sqrt {\tan x} } \right)}}{{\sin 2x}} - \dfrac{y}{{\sin 2x}} = 0 \\
    \\
  \left( {\dfrac{{dy}}{{dx}} - \sqrt {\tan x} } \right) - \dfrac{y}{{\sin 2x}} = 0 \\
    \\
    \\
$
Now, taking $\sqrt {\tan x} $ to the other side of the equation we will get the same value in positive making our following equation as:

$\dfrac{{dy}}{{dx}} - \dfrac{y}{{\sin 2x}} = \sqrt {\tan x} $ $ \to $ equation 1

As we know that we always find the integrating factor of the first order differential equations and the equation 1 written above is the first order differential equation, we will find its integrating factor in order to solve the solution further-
Integrating Factor = ${e^{\int {pdx} }}$
And the value of p from the above equation is $p = - \dfrac{1}{{\sin 2x}}$
Thus, Integrating Factor = ${e^{\int {pdx} }}$= ${e^{\int { - \dfrac{1}{{\sin 2x}}dx} }}$
Now, as we know that $\sin 2x = 2\sin x\cos x$, we will put this formula into ${e^{\int { - \dfrac{1}{{\sin 2x}}dx} }}$and further diving both numerator and denominator by ${\cos ^2}x$ we will make the solution simpler by cancelling $\cos x$ by $\cos x$ in the denominator giving us $2\dfrac{{\sin x}}{{\cos x}}$ we get:
${e^{\int { - \dfrac{1}{{2\sin x\cos x}}dx} }}$
Diving both numerator and denominator by ${\cos ^2}x$ we get
${e^{\int { - \dfrac{{\dfrac{1}{{{{\cos }^2}x}}}}{{\dfrac{{2\sin x\cos x}}{{{{\cos }^2}x}}}}dx} }}$
In the numerator we get ${\sec ^2}x$ as $\dfrac{1}{{{{\cos }^2}x}}$ = ${\sec ^2}x$ and in the denominator we will get $2\dfrac{{\sin x}}{{\cos x}}$ by cancelling $\cos x$ by $\cos x$. This will further give us $2\tan x$ in the denominator because $2\dfrac{{\sin x}}{{\cos x}} = 2\tan x$. We will get the equation as:

IF=${e^{\int { - \dfrac{{{{\sec }^2}x}}{{2\tan x}}dx} }}$

Now, let $\tan x = t$
Then ${\sec ^2}xdx = dt$
This will make the integrating factor as:
${e^{ - \int {\dfrac{{dt}}{{2t}}} }}$
solving further we get:
$
  {e^{ - \dfrac{1}{2}\ln t}} \\
   = {e^{\ln {{\left( t \right)}^{ - \dfrac{1}{2}}}}} \\
   = \dfrac{1}{{\sqrt t }} \\
$
So, IF of the equation is $\dfrac{1}{{\sqrt t }}$. Putting the value of $t = \tan x$ we will get,

=$\dfrac{1}{{\sqrt {\tan x} }}$

Solution is-
 $
  y.\dfrac{1}{{\sqrt {\tan x} }} = \int {\dfrac{1}{{\sqrt {\tan x} }}} .\sqrt {\tan x} dx \\
    \\
  \dfrac{y}{{\sqrt {\tan x} }} = x + c \\
 $

Since $\dfrac{1}{{\sqrt {\tan x} }} = \sqrt {\cot x} $, we get

$y\sqrt {\cot x} = x + c$

Note: In such questions, always try to make the given equation into first order differential equation so that the integral factor could be found and further formulas could be implied into the equation. It must look lengthy in the starting but is easy to understand.