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

Hint: 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.