Question

The general solution of $ \left( {2x - y + 1} \right)dx + \left( {2y - x + 1} \right)dy = 0 $ is
A) $ {x^2} + {y^2} + xy - x + y = c $
B) $ {x^2} + {y^2} - xy + x + y = c $
C) $ {x^2} - {y^2} + 2xy - x + y = c $
D) $ {x^2} + {y^2} - 2xy + x - y = c $

Answer 1

Hint: The given equation is a differential equation as the differentiation of the respective variables are present and represented as dx and dy. The general solution of this equation will be the equation constituted by only the variables and not their differentials. We can use the value of their integrals that are opposite to differentials so as to eliminate them and obtain the required general solution of the given equation:
Formulas to be used:
\[\dfrac{d}{{dx}}\left( {u.v} \right) = udv + vdu\]
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}dx\]

Complete step by step solution:
The given equation is:
 $ \left( {2x - y + 1} \right)dx + \left( {2y - x + 1} \right)dy = 0 $
We can see the presence of the terms dx and dy in this equation, this ‘d’ represents differentiation of the respective variables and thus the equation is known as differential equation.
To find its general solution we need to find an equation with the variables without their respective differentials.
By opening the brackets of the given equation we get:
\[\Rightarrow 2xdx - ydx + dx + 2ydy - xdy + dy......(1)\]
Here, we can observe that if we group together certain quantities, we can remove the sign of differentiation using their integrals..
 $
  i)\dfrac{d}{{dx}}\left( {x.y} \right) = xdy + ydx \\
\Rightarrow \left( {\because \dfrac{d}{{dx}}\left( {u.v} \right) = udv + vdu} \right) \\
 \Rightarrow \int {xdy + ydx = xy} \;
  $

\[
  ii)\dfrac{d}{{dx}}\left( {{x^2}} \right) = 2xdx;\dfrac{d}{{dx}}\left( {{y^2}} \right) = 2ydy \\
  \left( {\because \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}dx} \right) \\
   \Rightarrow \int {2xdx} = {x^2} \\
   \Rightarrow \int {2ydy} = {y^2} ;
 \]

 $
  iii)\dfrac{d}{{dx}}\left( x \right) = dx;\dfrac{d}{{dx}}\left( y \right) = dy \\
   \Rightarrow \int {dx = x} \\
   \Rightarrow \int {dy = y} \;
  $
Now, equation (1) can be written to make it compatible with above integration as:
\[2xdx + 2ydy - \left( {ydx + xdy} \right) + dx + dy = 0\]
Integrating both the sides, we get:
 $ \Rightarrow {x^2} + {y^2} - xy + x + y = c $ where c is any arbitrary constant.
Therefore, the general solution of $ \left( {2x - y + 1} \right)dx + \left( {2y - x + 1} \right)dy = 0 $ is $ {x^2} + {y^2} - xy + x + y = c $
So, the correct answer is “ $ {x^2} + {y^2} - xy + x + y = c $ ”.

Note: Differentiation and integration are opposite to each other. Thus, for the removal of differentiation signs we use their integrals. For example:
If differentiation of $ {x^2} $ is given by $ \dfrac{d}{{dx}}\left( {{x^2}} \right) = 2xdx $ , then the integration of $ 2xdx $ will be equal to $ {x^2} $ give as $ \int {2xdx = {x^2}} $
The general formula for both differentiation and integration respectively are:
For differentiation it is $ \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} $ and for integration is $ \int {{x^n} = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $