Question

If \[y(x)\]satisfies the differential equation \[y' - y\tan x = 2x\sec x\] and \[y(0) = 0\], then:
(This question has multiple correct options)
A.\[y\left( {\dfrac{\pi }{4}} \right) = \dfrac{{{\pi ^2}}}{{8\sqrt 2 }}\]
B. \[y'\left( {\dfrac{\pi }{4}} \right) = \dfrac{{{\pi ^2}}}{{18}}\]
C. \[y\left( {\dfrac{\pi }{3}} \right) = \dfrac{{{\pi ^2}}}{9}\]
D. \[y'\left( {\dfrac{\pi }{3}} \right) = \dfrac{{4\pi }}{3} + \dfrac{{2{\pi ^2}}}{{3\sqrt 3 }}\]

Answer 1

Hint: We solve for the solution of differential equation by comparing the equation to general from 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.
* Integrating factor is given by the formula \[I.F = {e^{\int {Pdx} }}\]
* \[\dfrac{d}{{dx}}\cot x = - \cos e{c^2}x\]
*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 \[y' - y\tan x = 2x\sec x\]
This equation is of the form \[\dfrac{{dy}}{{dx}} + P(x)y = Q(x)\], where \[P(x) = - \tan x;Q(x) = 2x\sec x\]
Now we find the integrating factor using the formula \[I.F = {e^{\int {Pdx} }}\]
\[\begin{gathered}
   \Rightarrow I.F = {e^{\int { - \tan xdx} }} \\
   \Rightarrow I.F = {e^{ - \int {\tan xdx} }} \\
\end{gathered} \]
We know that \[\int {\tan x = - \log \left| {\cos x} \right| + C} \]
Therefore, \[ - \int {\tan x = \log \left| {\cos x} \right|} + c\]
Here we ignore the constant value as we are taking power of exponential. Substitute the value of integration in the power.
\[ \Rightarrow I.F = {e^{\log (\cos x)}}\]
We know that log and exponential cancel each other
\[ \Rightarrow I.F = \cos x\]
We can write solution of the given differential equation as
\[ \Rightarrow y.\cos x = \int {2x\sec x.\cos xdx} + C\]
Substitute the value of \[\sec x = \dfrac{1}{{\cos x}}\]
\[ \Rightarrow y.\cos x = \int {2x\dfrac{1}{{\cos x}}.\cos xdx} + C\]
Cancel same factors
\[ \Rightarrow y.\cos x = \int {2xdx} + C\]
\[ \Rightarrow y.\cos x = \dfrac{{2{x^2}}}{2} + C\]
Cancel same factors
\[ \Rightarrow y.\cos x = {x^2} + C\] … (1)
Now we apply the given condition of particularity i.e. \[y(0) = 0\]
Put \[x = 0\]then we get \[y = 0\]
\[ \Rightarrow 0.\cos 0 = 0 + C\]
\[ \Rightarrow C = 0\]
So, equation (1) becomes \[y.\cos x = {x^2}\] … (2)
Differentiate both sides of the equation
\[ \Rightarrow \dfrac{d}{{dx}}\left( {y.\cos x} \right) = \dfrac{d}{{dx}}\left( {{x^2}} \right)\]
Use product rule in LHS
\[ \Rightarrow \dfrac{{dy}}{{dx}}\cos x + y( - \sin x) = 2x\]
\[ \Rightarrow \dfrac{{dy}}{{dx}}\cos x - y\sin x = 2x\]
Shift all values to RHS
\[ \Rightarrow \dfrac{{dy}}{{dx}}\cos x = 2x + y\sin x\]
\[ \Rightarrow y'\cos x = 2x + y\sin x\] … (3)
Now we calculate the value of ‘y’ and derivative of ‘y’ when \[x = \dfrac{\pi }{4}\]or \[x = \dfrac{\pi }{3}\]
When \[x = \dfrac{\pi }{4}\]:
\[ \Rightarrow y.\cos \dfrac{\pi }{4} = {\dfrac{\pi }{{16}}^2}\]
Put value of \[\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
\[ \Rightarrow y.\dfrac{1}{{\sqrt 2 }} = {\dfrac{\pi }{{16}}^2}\]
Cross multiply values from LHS to RHS
\[ \Rightarrow y = {\dfrac{{\sqrt 2 \pi }}{{16}}^2}\]
Cancel same factors from numerator and denominator
\[ \Rightarrow y = {\dfrac{\pi }{{8\sqrt 2 }}^2}\]
This is option A.

So, option A is correct.

Note: Students are advised to use differentiation and integration of common trigonometric functions directly. Also, use the values from the table if there is confusion in values of trigonometric functions.
Also, cancel the values from under root so as to obtain a matching answer from the options.

Angles (in degrees)	${0^ \circ }$	${30^ \circ }$	${45^ \circ }$	${60^ \circ }$	${90^ \circ }$
sin	0	$\dfrac{1}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{{\sqrt 3 }}{2}$	$1$
cos	1	$\dfrac{{\sqrt 3 }}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{1}{2}$	0
tan	0	$\dfrac{1}{{\sqrt 3 }}$	1	$\sqrt 3 $	Not defined
cosec	Not defined	2	\[\sqrt 2 \]	\[\dfrac{2}{{\sqrt 3 }}\]	1
sec	1	\[\dfrac{2}{{\sqrt 3 }}\]	\[\sqrt 2 \]	2	Not defined
cot	Not defined	$\sqrt 3 $	1	\[\dfrac{1}{{\sqrt 3 }}\]	0