Question

If $\dfrac{{dy}}{{dx}} + y\tan x = \sin 2x\;$ and $y\left( 0 \right) = 1$, then $y\left( \pi \right)$ is equal to
(A) 1
(B) −1
(C) −5
(D) 5

Answer 1

Here we will use the method of integrating factor t solve the given differential equation and the then use the given value to find the value of constant of integration and then finally find the value of $y\left( \pi \right)$
The method of integrating factor is:-
If a differential equation is of the form: $\dfrac{{dy}}{{dx}} + y.P\left( x \right) = Q\left( x \right)$
Then the integrating factor is given by:-
$I.F. = {e^{\int {p\left( x \right)dx} }}$
And the solution of the equation is given by:-
$y \times I.F. = \int {\left( {Q\left( x \right) \times I.F} \right)dx} + C$

Complete step-by-step answer:
The given differential equation is:-
$\dfrac{{dy}}{{dx}} + y\tan x = \sin 2x\;$
Since it is of the form:
$\dfrac{{dy}}{{dx}} + y.P\left( x \right) = Q\left( x \right)$
On comparing we get:-

$$p\left( x \right) = \tan x \\\ Q\left( x \right) = \sin 2x \\\$$

Hence we will use the method of integrating factor to solve this equation.
The integrating factor is given by:-
$I.F. = {e^{\int {p\left( x \right)dx} }}$
Hence in this equation the integrating factor is:-
$I.F. = {e^{\int {\tan xdx} }}$
Now we know that:-
$\int {\tan xdx = \log \left( {\sec x} \right)} + C$
Hence putting the value we get:-
$I.F. = {e^{\log \sec x}}$
Simplifying it we get:-
$I.F. = \sec x$
Now the solution of the equation is given by:-
$y \times I.F. = \int {\left( {Q\left( x \right) \times I.F} \right)dx} + C$
Hence putting the respective values we get:-
$y \times \sec x = \int {\left( {\sin 2x \times \sec x} \right)dx} + C$
Now we know that:-

$$\sin 2x = 2\sin x\cos x \\\ \sec x = \dfrac{1}{{\cos x}} \\\$$

Hence substituting these values we get:-
$y \times \sec x = \int {\left( {2\sin x\cos x \times \dfrac{1}{{\cos x}}} \right)dx} + C$
Simplifying it further we get:-

$$y \times \sec x = \int {\left( {2\sin x} \right)dx} + C \\\ \Rightarrow y \times \sec x = 2\int {\left( {\sin x} \right)dx} + C \\\$$

Now we know that:-
$\int {\sin xdx = - \cos x + C}$
Hence putting this value in above equation we get:-
$y \times \sec x = 2\left( { - \cos x} \right) + C$
We know that:-
$\sec x = \dfrac{1}{{\cos x}}$
Putting the value we get:-
$\dfrac{y}{{\cos x}} = 2\left( { - \cos x} \right) + C$
Simplifying it we get:-
$y = - 2{\cos ^2}x + C\cos x$……………………………….(1)
Now putting in the given value i.e. $y\left( 0 \right) = 1$
We get:-
$1 = - 2{\cos ^2}\left( 0 \right) + C\cos \left( 0 \right)$
We know that:-
$\cos 0 = 1$
Hence substituting this value we get:-
$1 = - 2\left( 1 \right) + C\left( 1 \right)$
Evaluating the value of C we get:-

$$C = 1 + 2 \\\ \Rightarrow C = 3 \\\$$

Putting this value in equation 1 we get:-
$y = - 2{\cos ^2}x + 3\cos x$…………………………..(2)
Now we have to find $y\left( \pi \right)$
Hence putting $x = \pi$in equation 2 we get:-
$y = - 2{\cos ^2}\left( \pi \right) + 3\cos \left( \pi \right)$
Now we know that:
$\cos \left( \pi \right) = - 1$
Putting this value we get:-
$y = - 2{\left( { - 1} \right)^2} + 3\left( { - 1} \right)$
Solving it further we get:-

$$y = - 2 - 3 \\\ \Rightarrow y = - 5 \\\$$

Hence option C is correct.

Note: In the questions of differential equations we have to observe which form is given and then solve it accordingly.
Students may make mistakes while evaluating the value of C so all the values should be substituted carefully.