Question

What is the solution of the differential equation \[dy - \sin x\sin ydx = 0\]?
A. \[{e^{\cos x}}\tan \dfrac{y}{2} = c\]
B. \[{e^{\cos x}}\tan y = c\]
C. \[\cos x\tan y = c\]
D. \[\cos x\sin y = c\]

Answer 1

Hint: The given differential equation consists of two variables that are x and y. First we will separate the variables of the given differential equation and integrate both sides to get required solution.

Formula Used: Integration formula of trigonometry identities:
\[\int {\sin xdx = - \cos x + c} \]
\[\int {\dfrac{1}{x}dx} = \log x + c\]
Half angle formula in trigonometry:
\[\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}\]
Quotient logarithm formula:
\[\log a - \log b = \log \dfrac{a}{b}\]

Complete step by step solution: Given differential equation is:
\[dy - \sin x\sin ydx = 0\]
The variables of the differential equation are x and y.
Now separates the variables of the equation:
\[ \Rightarrow dy = \sin x\sin ydx\]
Divide both sides by \[\sin y\]
\[ \Rightarrow \dfrac{{dy}}{{\sin y}} = \sin xdx\]
Apply half angle formula on \[\sin y\]:
\[ \Rightarrow \dfrac{{dy}}{{2\sin \dfrac{y}{2}\cos \dfrac{y}{2}}} = \sin xdx\]
Taking integration on both sides:
\[ \Rightarrow \int {\dfrac{{dy}}{{2\sin \dfrac{y}{2}\cos \dfrac{y}{2}}}} = \int {\sin xdx} \] ….(i)
Divide \[{\cos ^2}\dfrac{y}{2}\] with the denominator and numerator of \[\int {\dfrac{{dy}}{{2\sin \dfrac{y}{2}\cos \dfrac{y}{2}}}} \]
\[ \Rightarrow \int {\dfrac{{{{\sec }^2}\dfrac{y}{2}dy}}{{2\tan \dfrac{y}{2}}}} = \int {\sin xdx} \]
Assume that, \[{I_1} = \int {\dfrac{{{{\sec }^2}\dfrac{y}{2}dy}}{{2\tan \dfrac{y}{2}}}} \]
Let \[\tan \dfrac{y}{2} = z\]
Different both sides
\[\dfrac{1}{2}{\sec ^2}\dfrac{y}{2}dy = dz\]
Substitute \[\dfrac{1}{2}{\sec ^2}\dfrac{y}{2}dy = dz\] and \[\tan \dfrac{y}{2} = z\] in \[{I_1} = \int {\dfrac{{{{\sec }^2}\dfrac{y}{2}dy}}{{2\tan \dfrac{y}{2}}}} \]
\[{I_1} = \int {\dfrac{{dz}}{z}} \]
Applying the formula \[\int {\dfrac{1}{x}dx} = \log x + c\]
\[{I_1} = \log z + {c_1}\]
Substitute the value of z:
\[{I_1} = \log \left( {\tan \dfrac{y}{2}} \right) + {c_1}\]
Now we will substitute \[\int {\dfrac{{{{\sec }^2}\dfrac{y}{2}dy}}{{2\tan \dfrac{y}{2}}}} = \log \left( {\tan \dfrac{y}{2}} \right) + {c_1}\] in equation (i)
\[ \Rightarrow \log \left( {\tan \dfrac{y}{2}} \right) = - \cos x + \log c\]
\[ \Rightarrow \log \left( {\tan \dfrac{y}{2}} \right) - \log c = - \cos x\]
Applying quotient rule:
\[ \Rightarrow \log \left( {\dfrac{{\tan \dfrac{y}{2}}}{c}} \right) = - \cos x\]
Applying the logarithm inverse formula :
\[ \Rightarrow \dfrac{{\tan \dfrac{y}{2}}}{c} = {e^{ - \cos x}}\]
Now simplify the above equation:
\[ \Rightarrow \dfrac{{\tan \dfrac{y}{2}}}{{{e^{ - \cos x}}}} = c\]
\[ \Rightarrow {e^{ - \cos x}}\tan \dfrac{y}{2} = c\]

Option ‘A’ is correct

Note: Students often do mistake to integrate \[\dfrac{{dy}}{{\sin y}} = \csc ydy\]. There are four formula to integrate \[\csc ydy\]. The integration formulas are \[\int {\csc xdx} = \left\{ {\begin{array}{*{20}{c}}{\log \left| {\csc x - \cot x} \right| + c}\\{ - \log \left| {\csc x + \cot x} \right| + c}\\{\dfrac{1}{2}\log \left| {\dfrac{{\cos x - 1}}{{\cos x + 1}}} \right| + c}\\{\log \left| {\tan \dfrac{x}{2}} \right| + c}\end{array}} \right.\]. If they used any of them except \[\int {\csc xdx} = \log \left( {\tan \dfrac{x}{2}} \right) + c\], then they will not get the correct answer.