Question

The solution of \[\dfrac{{dy}}{{dx}} = \sin (x + y) + \cos (x + y)\]is:
A. \[\log \left[ {1 + \tan \left( {\dfrac{{x + y}}{2}} \right)} \right] + c = 0\]
B. \[\log \left[ {1 + \tan \left( {\dfrac{{x + y}}{2}} \right)} \right] = x + c\]
C. \[\log \left[ {1 - \tan \left( {\dfrac{{x + y}}{2}} \right)} \right] = x + c\]
D. None of these.

Answer 1

Hint:
Some formulae of integration that can be useful in solving such questions:
1) \[\int {dx = x + C} \]
2) \[\int {\dfrac{{dx}}{x} = \log |x| + C} \]
Where, C= constant of integration.
Variable separable method is used for solving differential equations in which the variables can be separated easily and can be integrable.

Complete step by step solution:
Given: \[\dfrac{{dy}}{{dx}} = \sin (x + y) + \cos (x + y)\]
Put \[x + y\]=t;
\[
   \Rightarrow 1 + \dfrac{{dy}}{{dx}} = \dfrac{{dt}}{{dx}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dt}}{{dx}} - 1 \\
 \]
Putting above values we get,
\[ \Rightarrow \dfrac{{dt}}{{dx}} - 1 = \sin t + \cos t\]
\[ \Rightarrow \dfrac{{dt}}{{dx}} = \sin t + \cos t + 1\]
Now using separation of variables we get;
\[ \Rightarrow \dfrac{{dt}}{{\sin t + \cos t + 1}} = dx\]
Simplifying the above equation we get;
\[
   \Rightarrow \dfrac{{dt}}{{\sin t + \cos t + 1}} = dx \\
   \Rightarrow \dfrac{{dt}}{{2\sin \left( {\dfrac{t}{2}} \right)\cos \left( {\dfrac{t}{2}} \right) + 2{{\cos }^2}\left( {\dfrac{t}{2}} \right)}} = dx........(u\sin g:1 + \cos 2x = 2{\cos ^2}x,\sin 2x = 2\sin x\cos x) \\
   \Rightarrow \dfrac{{dt}}{{2\cos \left( {\dfrac{t}{2}} \right)\left( {\sin \left( {\dfrac{t}{2}} \right) + \cos \left( {\dfrac{t}{2}} \right)} \right)}} = dx \\
   \Rightarrow \dfrac{1}{2}\dfrac{{\sec \left( {\dfrac{t}{2}} \right)dt}}{{\left( {\sin \left( {\dfrac{t}{2}} \right) + \cos \left( {\dfrac{t}{2}} \right)} \right)}} = dx \\
 \]
Dividing each term by \[\cos \left( {\dfrac{t}{2}} \right)\]:
\[ \Rightarrow \dfrac{1}{2}\dfrac{{{{\sec }^2}\left( {\dfrac{t}{2}} \right)}}{{\left( {1 + \tan \left( {\dfrac{t}{2}} \right)} \right)}}dt = dx.......EQ:01\]
To solve above equation put \[1 + \tan \left( {\dfrac{t}{2}} \right) = z\]
\[
   \Rightarrow {\sec ^2}\left( {\dfrac{t}{2}} \right) \times \dfrac{1}{2}dt = dz \\
   \Rightarrow {\sec ^2}\left( {\dfrac{t}{2}} \right)dt = 2dz \\
 \]
Putting \[{\sec ^2}\left( {\dfrac{t}{2}} \right)dt = 2dz\]in EQ:01,
\[
   \Rightarrow \dfrac{1}{2}(2)\dfrac{{dz}}{{\left( z \right)}} = dx \\
   \Rightarrow \dfrac{{dz}}{{\left( z \right)}} = dx \\
 \]
On integrating we get;
\[
   \Rightarrow \int {\dfrac{{dz}}{{\left( z \right)}}} = \int {dx} \\
   \Rightarrow \log |z| = x + c \\
 \]
Put back the value-: \[z = 1 + \tan \left( {\dfrac{t}{2}} \right)\].
\[ \Rightarrow \log |1 + \tan \left( {\dfrac{t}{2}} \right)| = x + c\]
Again put \[x + y\]=t in above equation;
\[ \Rightarrow \log |1 + \tan \left( {\dfrac{{x + y}}{2}} \right)| = x + c\]

Option (B) is correct.

Note:
Student’s have to check whether the variables are in simpler form or complex form. If they are in complex form by substituting make them in simpler form. After that you can use methods like variable separation to solve the differential equation.