Question

What is the solution of the differential equation \[\dfrac{{dy}}{{dx}}\tan y = \sin \left( {x + y} \right) + \sin \left( {x - y} \right)\]?
A. \[\sec y + 2\cos x = c\]
B. \[\sec y - 2\cos x = c\]
C. \[\cos y - 2\sin x = c\]
D. \[\tan y - 2\sec y = c\]
E. \[\sec y + 2\sin x = c\]

Answer 1

Hint: First we will apply sum of two sin angles to simplify the right hand side expression of the given differential equation. Then we will separate the variables of the differential equation and integrating both sides to solve the given differential equation.

Formula Used: Integration of trigonometry ratios:
\[\int {\sin \theta d\theta } = - \cos \theta + c\]
\[\int {\sec \theta \tan \theta d\theta } = \sec \theta + c\]
Sum of sin angles:
\[\sin A + \sin B = 2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}\]

Complete step by step solution: The given differential equation is
\[\dfrac{{dy}}{{dx}}\tan y = \sin \left( {x + y} \right) + \sin \left( {x - y} \right)\]’
Applying sum of sin angles in the right side expression:
\[ \Rightarrow \dfrac{{dy}}{{dx}}\tan y = 2\sin \left( {\dfrac{{x + y + x - y}}{2}} \right)\cos \left( {\dfrac{{x + y - x + y}}{2}} \right)\]
\[ \Rightarrow \dfrac{{dy}}{{dx}}\tan y = 2\sin x\cos y\]
Now we will separate the variables of the differential equation:
\[ \Rightarrow \dfrac{{\tan y}}{{\cos y}}dy = 2\sin xdx\]
Rewrite the above equation:
\[ \Rightarrow \tan y\sec ydy = 2\sin xdx\]
Now integrating both sides:
\[ \Rightarrow \int {\tan y\sec ydy} = 2\int {\sin xdx} \]
Applying the formula of the integration:
\[ \Rightarrow \sec y = - 2\cos x + c\]
\[ \Rightarrow \sec y + 2\cos x = c\]

Option ‘A’ is correct

Additional Information: The constant that we added with the solution of the differential equation is known as integrating constant.

Note: Students often do mistake to integrate \[\sin x\]. They forgot to put negative sign of the integration of \[\sin x\]. They used \[\int {\sin \theta d\theta } = \cos \theta + c\]. The correct formula is \[\int {\sin \theta d\theta } = - \cos \theta + c\]