Question

If $ \left( {dx + dy} \right) = \left( {x + y} \right)\left( {dx - dy} \right) $ , then find the value of $ \log \left( {x + y} \right) $ ?

Answer 1

Hint: In the given question, we are given a differential equation. So, we have to solve the given differential equation using methods of integration. Then, we will find the value of the expression provided to us in the question itself. We first convert the given differential equation into a form which is easier to solve compared to the one given in the problem itself.

Complete step-by-step answer:
The given question requires us to solve a differential equation $ \left( {dx + dy} \right) = \left( {x + y} \right)\left( {dx - dy} \right) $ using methods of integration and then find the value of expression $ \log \left( {x + y} \right) $ .
First of all, we have to classify the type of differential equation.
The given differential equation is not of variable separable form as we cannot separate the variables in the differential equation.
Also, the differential equation is also not a homogeneous equation as all the terms don’t have the same degree of variables.
We will have converted the given differential equation into a form which is easier to solve compared to the one given in the problem itself.
First, we shift all the differential terms to the left side of the equation, we get,
 $ \Rightarrow \dfrac{{\left( {dx + dy} \right)}}{{\left( {dx - dy} \right)}} = \left( {x + y} \right) $
Now, using the componendo and dividend concept, we get,
\[ \Rightarrow \dfrac{{\left( {dx + dy} \right) + \left( {dx - dy} \right)}}{{\left( {dx + dy} \right) - \left( {dx - dy} \right)}} = \dfrac{{\left( {x + y} \right) + 1}}{{\left( {x + y} \right) - 1}}\]
Simplifying the expression further, we get,
\[ \Rightarrow \dfrac{{2dx}}{{2dy}} = \dfrac{{\left( {x + y} \right) + 1}}{{\left( {x + y} \right) - 1}}\]
Cancelling the common factors in numerator and denominator, we get the value of $ \dfrac{{dy}}{{dx}} $ as,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {x + y} \right) - 1}}{{\left( {x + y} \right) + 1}}\]
So, we substitute $ x + y = t $ . Then, differentiating both sides of this equation with respect to x, we get,
 $ \Rightarrow 1 + \dfrac{{dy}}{{dx}} = \dfrac{{dt}}{{dx}} $
Isolating the term $ \dfrac{{dy}}{{dx}} $ in order to find its value, we get,
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dt}}{{dx}} - 1 $
Now substituting the value of $ \dfrac{{dy}}{{dx}} $ as $ \left( {\dfrac{{dt}}{{dx}} - 1} \right) $ and value of $ x + y $ as t in the differential equation \[\dfrac{{dy}}{{dx}} = \dfrac{{\left( {x + y} \right) - 1}}{{\left( {x + y} \right) + 1}}\], we get,
\[ \Rightarrow \dfrac{{dt}}{{dx}} - 1 = \dfrac{{t - 1}}{{t + 1}}\]
Shifting all the terms to right side of the equation except the differential, we get,
\[ \Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{{t - 1}}{{t + 1}} + 1\]
Simplifying the expression, we get,
\[ \Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{{t - 1 + t + 1}}{{t + 1}}\]
\[ \Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{{2t}}{{t + 1}}\]
\[ \Rightarrow \dfrac{{t + 1}}{{2t}}dt = dx\]
Integrating both sides of the equation, we get,
\[ \Rightarrow \int {\dfrac{{t + 1}}{{2t}}dt} = \int {dx} \]
\[ \Rightarrow \int {\left( {\dfrac{1}{2} + \dfrac{1}{{2t}}} \right)dt} = x\]
\[ \Rightarrow \left( {\dfrac{t}{2} + \dfrac{1}{2}\ln \left| t \right|} \right) = x + k\]
Substituting back the value of t as $ \left( {x + y} \right) $ , we get,
\[ \Rightarrow \dfrac{{x + y}}{2} + \dfrac{1}{2}\ln \left| {x + y} \right| = x + k\], where k is an arbitrary constant
Shifting all the terms to right side of the equation except for $ \ln \left( {x + y} \right) $ , we get,
\[ \Rightarrow \dfrac{1}{2}\ln \left| {x + y} \right| = x - \left( {\dfrac{{x + y}}{2}} \right) + k\]
\[ \Rightarrow \dfrac{1}{2}\ln \left| {x + y} \right| = \left( {\dfrac{{x - y}}{2}} \right) + k\]
Multiplying both sides of the equation by $ 2 $ , we get,
\[ \Rightarrow \ln \left| {x + y} \right| = x - y + 2k\]
\[ \Rightarrow \ln \left| {x + y} \right| = x - y + c\], where $ c = 2k $ , is an arbitrary constant.
Therefore, the value of \[\ln \left| {x + y} \right|\] is \[\left( {x - y + c} \right)\] given the differential equation $ \left( {dx + dy} \right) = \left( {x + y} \right)\left( {dx - dy} \right) $ .
So, the correct answer is “\[\left( {x - y + c} \right)\] ”.

Note: The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the constant. Exact value of the arbitrary constant can be calculated only if we are given a point lying on the function.