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

Hint: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.