Question

The solution to the equation $\dfrac{{dy}}{{dx}} = \sin \left( {x + y} \right) + \cos \left( {x + y} \right)$ 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

First we’ll substitute the dependent variable with a new variable then will proceed to solve for that variable, here we’ll substitute (x+y) with a new variable, then on differentiating with-respect-to independent variable we’ll also substitute the derivative of the dependent variable, then will further proceed.

Complete step by step answer:

Given data: $\dfrac{{dy}}{{dx}} = \sin \left( {x + y} \right) + \cos \left( {x + y} \right).............(i)$
Let us assume that, $x + y = t$
On differentiating with respect to ‘x’,
i.e. $1 + \dfrac{{dy}}{{dx}} = \dfrac{{dt}}{{dx}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dt}}{{dx}} - 1$
On substituting the value of$\dfrac{{dy}}{{dx}}$and $'x + y'$ in equation (i), we get,
$\dfrac{{dt}}{{dx}} - 1 = \sin t + \cos t$
$\Rightarrow \dfrac{{dt}}{{dx}} = 1 + \sin t + \cos t$
Using the double-angle formula of sin(A) and cos(A)
i.e. $\sin (2A) = 2\sin A\cos A$
And, $\cos (2A) = 2{\cos ^2}A - 1$
We’ll get,
$\Rightarrow \dfrac{{dt}}{{dx}} = 1 + 2\sin \left( {\dfrac{t}{2}} \right)\cos \left( {\dfrac{t}{2}} \right) + 2{\cos ^2}\left( {\dfrac{t}{2}} \right) - 1$
$\Rightarrow \dfrac{{dt}}{{dx}} = 2\sin \left( {\dfrac{t}{2}} \right)\cos \left( {\dfrac{t}{2}} \right) + 2{\cos ^2}\left( {\dfrac{t}{2}} \right)$
Now solving for ‘t’ and ‘x’, we’ll use the variable separation method
$\Rightarrow \dfrac{{dt}}{{2\sin \left( {\dfrac{t}{2}} \right)\cos \left( {\dfrac{t}{2}} \right) + 2{{\cos }^2}\left( {\dfrac{t}{2}} \right)}} = dx$
Multiplying and dividing left side by ${\sec ^2}\left( {\dfrac{t}{2}} \right)$ , we get,
$\Rightarrow \dfrac{{{{\sec }^2}\left( {\dfrac{t}{2}} \right)dt}}{{2{{\sec }^2}\left( {\dfrac{t}{2}} \right)\sin \left( {\dfrac{t}{2}} \right)\cos \left( {\dfrac{t}{2}} \right) + 2{{\sec }^2}\left( {\dfrac{t}{2}} \right){{\cos }^2}\left( {\dfrac{t}{2}} \right)}} = dx$
On using, ${\sec ^2}A = \dfrac{1}{{{{\cos }^2}A}}$ in the denominator, we get,
$\Rightarrow \dfrac{{{{\sec }^2}\left( {\dfrac{t}{2}} \right)dt}}{{2\dfrac{{\sin \left( {\dfrac{t}{2}} \right)\cos \left( {\dfrac{t}{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{t}{2}} \right)}} + 2\dfrac{{{{\cos }^2}\left( {\dfrac{t}{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{t}{2}} \right)}}}} = dx$
$\Rightarrow \dfrac{{{{\sec }^2}\left( {\dfrac{t}{2}} \right)dt}}{{2\dfrac{{\sin \left( {\dfrac{t}{2}} \right)}}{{\cos \left( {\dfrac{t}{2}} \right)}} + 2}} = dx$
Now, using$\dfrac{{\sin A}}{{\cos A}} = \tan A$, we get,
$\Rightarrow \dfrac{{{{\sec }^2}\left( {\dfrac{t}{2}} \right)dt}}{{2\tan \left( {\dfrac{t}{2}} \right) + 2}} = dx$
$\Rightarrow \dfrac{{{{\sec }^2}\left( {\dfrac{t}{2}} \right)dt}}{{2\left[ {\tan \left( {\dfrac{t}{2}} \right) + 1} \right]}} = dx........(ii)$
Now, let $p = 1 + \tan \left( {\dfrac{t}{2}} \right)$
On differentiating with respect to ‘t’, we get,
$\Rightarrow \dfrac{{dp}}{{dt}} = \dfrac{1}{2}{\sec ^2}\left( {\dfrac{t}{2}} \right)$
$\Rightarrow dp = \dfrac{1}{2}{\sec ^2}\left( {\dfrac{t}{2}} \right)dt$
Substituting the value of $1 + \tan \left( {\dfrac{t}{2}} \right)$ and $\dfrac{1}{2}{\sec ^2}\left( {\dfrac{t}{2}} \right)$ in equation (ii), we get,
$\Rightarrow \dfrac{{dp}}{p} = dx$
Now taking the integral of both the sides,
$\Rightarrow \smallint \dfrac{{dp}}{p} = \smallint dx$
Using $\smallint \dfrac{{dz}}{z} = \log z + c$ and $\smallint dx = x + c$, we get,
$\Rightarrow \log p = x + C$
Substituting the value of ‘p’, we get,
$\Rightarrow \log \left( {1 + \tan \left( {\dfrac{t}{2}} \right)} \right) = x + C$
Now, substituting the value of ‘t’, we get,
$\therefore \log \left( {1 + \tan \left( {\dfrac{{x + y}}{2}} \right)} \right) = x + C$
Therefore, option (B) is correct

Note: In the above solution we used the variable separable method to solve the particular differential equation.
Generally, variable separable method is used when we can rewrite the equation such that on either side of the equation there are terms of only one variable,
$f(x)g(y)dx = g(x)f(y)dy$
Then on separating the similar variable terms,
$\Rightarrow \dfrac{{f(x)}}{{g(x)}}dx = \dfrac{{f(y)}}{{g(y)}}dy$, than we can simplify easily.