Question

The general solution of the differential equation $\sin 2x(\dfrac{{dy}}{{dx}} - \sqrt {\tan x} ) - y = 0$ is
A. $y\sqrt {\tan x} = x + c$
B.$y\sqrt {\cot x} = \tan x + c$
C.$t\sqrt {\tan x} = \cot x + c$
D.$y\sqrt {\cot x} = x + c$

Answer 1

We solve for the solution of differential equation by comparing the equation to general form of differential equation which will give us the values of $P(x),Q(x)$ and then we find the integrating factor using the formula and multiply both sides of the equation by integrating factor and then integrate both sides.

Formula used:
a) Integrating factor is given by the formula $I.F = {e^{\int {Pdx} }}$
b) $\dfrac{d}{{dx}}\cot x = - \cos e{c^2}x$
c) Chain rule of differentiation says $\dfrac{d}{{dx}}f(g(x)) = f'(g(x)) \times g'(x)$ where $f'(g(x))$ is differentiation of f with respect to x and $g'(x)$ is differentiation of g with respect to x.

Complete step-by-step answer:
We are given the differential equation $\sin 2x(\dfrac{{dy}}{{dx}} - \sqrt {\tan x} ) - y = 0$
We shift the variable y to one side of the equation
$\Rightarrow \sin 2x(\dfrac{{dy}}{{dx}} - \sqrt {\tan x} ) = y$
Divide both sides of the equation by $\sin 2x$
$\Rightarrow \dfrac{{\sin 2x(\dfrac{{dy}}{{dx}} - \sqrt {\tan x} )}}{{\sin 2x}} = \dfrac{y}{{\sin 2x}}$
Cancel out same terms from numerator and denominator
$\Rightarrow \dfrac{{dy}}{{dx}} - \sqrt {\tan x} = \dfrac{y}{{\sin 2x}}$
Shift the term $\dfrac{y}{{\sin 2x}}$ to LHS of the equation and the term $\sqrt {\tan x}$ to RHS of the equation.
$\Rightarrow \dfrac{{dy}}{{dx}} - \dfrac{y}{{\sin 2x}} = \sqrt {\tan x} ………. (1)$
This equation is of the form $\dfrac{{dy}}{{dx}} + P(x)y = Q(x)$, where $P(x) = \dfrac{{ - 1}}{{\sin 2x}} = - \cos ec2x,Q(x) = \sqrt {\tan x}$ {Since we know $\dfrac{1}{{\sin x}} = \cos ecx$}
Now we find the integrating factor using the formula $I.F = {e^{\int {Pdx} }}$

$$\Rightarrow I.F = {e^{\int { - \cos ec2xdx} }} \\\ \Rightarrow I.F = {e^{ - \int {\cos ec2xdx} }} \\\$$

We know that $\int {\cos ecx = \log \left| {\tan \dfrac{x}{2}} \right|} + c$
Therefore, $\int {\cos ec2x = \dfrac{1}{2}\log \left| {\tan \dfrac{{2x}}{2}} \right|} + c = \dfrac{1}{2}\log \left| {\tan x} \right| + c$ { integration of $2x$ will give $\dfrac{1}{2}$}
Here we ignore the constant value as we are taking power of exponential. Substitute the value of integration in the power.
$\Rightarrow I.F = {e^{ - \dfrac{1}{2}\log (\tan x)}}$
Now we know that $m\log x = \log {x^m}$
$\Rightarrow I.F = {e^{\log {{(\tan x)}^{ - \dfrac{1}{2}}}}}$
Also, we know that ${(\tan x)^{ - \dfrac{1}{2}}} = \dfrac{1}{{\sqrt {\tan x} }} = \sqrt {\cot x}$
$\Rightarrow I.F = {e^{\log (\sqrt {\cot x} )}}$
We know that log and exponential cancel each other
$\Rightarrow I.F = \sqrt {\cot x}$
Now we multiply both sides of the equation (1) by $I.F = \sqrt {\cot x}$
$\Rightarrow \sqrt {\cot x} \dfrac{{dy}}{{dx}} - \dfrac{{y\sqrt {\cot x} }}{{\sin 2x}} = \sqrt {\cot x} \sqrt {\tan x} …….. (2)$
Now we can write LHS of the equation as
$\dfrac{d}{{dx}}(y\sqrt {\cot x} )$ because when we use chain rule i.e. $\dfrac{d}{{dx}}f(g(x)) = f'(g(x)) \times g'(x)$ to find differentiation of $(y\sqrt {\cot x} )$ we see
$\Rightarrow \dfrac{d}{{dx}}(y\sqrt {\cot x} ) = \dfrac{{dy}}{{dx}}(\sqrt {\cot x} ) + y\dfrac{d}{{dx}}(\sqrt {\cot x} )……... (3)$
Here we have to solve for $\dfrac{d}{{dx}}(\sqrt {\cot x} )$
$\Rightarrow \dfrac{d}{{dx}}(\sqrt {\cot x} ) = \dfrac{d}{{dx}}{(\cot x)^{\dfrac{1}{2}}}$
Using the product rule of differentiation, and $\dfrac{d}{{dx}}(\cot x) = - \cos e{c^2}x$

$$\Rightarrow \dfrac{d}{{dx}}(\sqrt {\cot x} ) = \dfrac{1}{2}{(\cot x)^{\dfrac{1}{2} - 1}} \times ( - \cos e{c^2}x) \\\ \Rightarrow \dfrac{d}{{dx}}(\sqrt {\cot x} ) = - \dfrac{1}{2}{(\cot x)^{ - \dfrac{1}{2}}} \times \cos e{c^2}x \\\$$

Writing $\cot {x^{ - \dfrac{1}{2}}} = \dfrac{1}{{\sqrt {\cot x} }}$
\Rightarrow \dfrac{d}{{dx}}(\sqrt {\cot x} ) = - \dfrac{1}{2}\left\\{ {\dfrac{1}{{\sqrt {\cot x} }}} \right\\} \times \cos e{c^2}x
Rationalize the term inside the bracket by multiplying both numerator and denominator by $\sqrt {\cot x}$

\Rightarrow \dfrac{d}{{dx}}(\sqrt {\cot x} ) = - \dfrac{1}{2}\left\\{ {\dfrac{1}{{\sqrt {\cot x} }} \times \dfrac{{\sqrt {\cot x} }}{{\sqrt {\cot x} }}} \right\\} \times \dfrac{1}{{{{\sin }^2}x}} \\\ \Rightarrow \dfrac{d}{{dx}}(\sqrt {\cot x} ) = - \dfrac{1}{2}\left\\{ {\dfrac{{\sqrt {\cot x} }}{{{{\left( {\sqrt {\cot x} } \right)}^2}}}} \right\\} \times \dfrac{1}{{{{\sin }^2}x}} \\\

Cancel square root by square power.
\Rightarrow \dfrac{d}{{dx}}(\sqrt {\cot x} ) = - \dfrac{1}{2}\left\\{ {\dfrac{{\sqrt {\cot x} }}{{\cot x}}} \right\\} \times \dfrac{1}{{{{\sin }^2}x}}
Write $\cot x = \dfrac{{\cos x}}{{\sin x}}$
$\Rightarrow \dfrac{d}{{dx}}(\sqrt {\cot x} ) = - \dfrac{1}{2}\dfrac{{\sin x\sqrt {\cot x} }}{{\cos x}} \times \dfrac{1}{{{{\sin }^2}x}}$
Cancel out same terms from numerator and denominator
$\Rightarrow \dfrac{d}{{dx}}(\sqrt {\cot x} ) = - \dfrac{1}{2}\dfrac{{\sqrt {\cot x} }}{{\cos x}} \times \dfrac{1}{{\sin x}}$
Write the value of $2\sin x\cos x = \sin 2x$ in the denominator.
$\Rightarrow \dfrac{d}{{dx}}(\sqrt {\cot x} ) = - \dfrac{{\sqrt {\cot x} }}{{\sin 2x}}$
Substitute the value in equation (3)
$\Rightarrow \dfrac{d}{{dx}}(y\sqrt {\cot x} ) = \dfrac{{dy}}{{dx}}\sqrt {\cot x} - \dfrac{{y\sqrt {\cot x} }}{{\sin 2x}}$ … (4)
Now from equation (4) we can write equation (2) as
$\Rightarrow \dfrac{d}{{dx}}y\sqrt {\cot x} = \sqrt {\cot x} \sqrt {\tan x}$
Since, $\sqrt {\cot x} = \dfrac{1}{{\sqrt {\tan x} }}$

$$\Rightarrow \dfrac{d}{{dx}}y\sqrt {\cot x} = \dfrac{1}{{\sqrt {\tan x} }}\sqrt {\tan x} \\\ \Rightarrow \dfrac{d}{{dx}}y\sqrt {\cot x} = 1 \\\$$

Shift dx to RHs of the equation
$\Rightarrow dy\sqrt {\cot x} = dx$
Now we integrate both sides of the equation (here $\sqrt {\cot x}$ is taken as constant on LHS)

$$\Rightarrow \int {\sqrt {\cot x} dy} = \int {dx} \\\ \Rightarrow y\sqrt {\cot x} = x + c \\\$$

So, the correct answer is “Option D”.

Note: Students are likely to make mistake in the part where we convert LHS as differentiation of $(y\sqrt {\cot x} )$ because many students don’t know the differentiation of $\cot x = - \cos e{c^2}x$, so they make it more complex. Students are advised to use differentiation and integration of common trigonometric functions directly.