Question

Solve the given equation \[\left( {x + y - 1} \right)dx + \left( {2x + 2y - 3} \right)dy = 0\]

Answer 1

Hint: This problem can be solved by using a substitution method. According to the substitution method, the given integral can be transformed into another form by changing the independent variable \[x{\text{ to }}t\]. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given equation is \[\left( {x + y - 1} \right)dx + \left( {2x + 2y - 3} \right)dy = 0\] which can be written as
\[
   \Rightarrow \left( {2x + 2y - 3} \right)dy = - \left( {x + y - 1} \right)dx \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \left( {x + y - 1} \right)}}{{2x + 2y - 3}} = \dfrac{{ - \left( {x + y} \right) + 1}}{{2\left( {x + y} \right) - 3}} \\
\]
Put \[x + y = t\] then, by differentiating it on both sides we have \[1 + \dfrac{{dy}}{{dx}} = \dfrac{{dt}}{{dx}}\] i.e., \[\dfrac{{dy}}{{dx}} = \dfrac{{dt}}{{dx}} - 1\]. So, we have
\[
  \dfrac{{dt}}{{dx}} - 1 = \dfrac{{ - t + 1}}{{2t - 3}} \\
  \dfrac{{dt}}{{dx}} = \dfrac{{1 - t}}{{2t - 3}} + 1 \\
  \dfrac{{dt}}{{dx}} = \dfrac{{1 - t + 2t - 3}}{{2t - 3}} \\
  \dfrac{{dt}}{{dx}} = \dfrac{{t - 2}}{{2t - 3}} \\
  \left( {\dfrac{{2t - 3}}{{t - 2}}} \right)dt = dx \\
\]
Adding and subtracting 1 on the numerator of right-hand side, we get
\[
  \left( {\dfrac{{2t - 4 + 1}}{{t - 2}}} \right)dt = dx \\
  \left( {\dfrac{{2\left( {t - 2} \right) + 1}}{{t - 2}}} \right)dt = dx \\
\]
 Splitting the terms, we have
\[
  \left( {\dfrac{{2\left( {t - 2} \right)}}{{t - 2}} + \dfrac{1}{{t - 2}}} \right)dt = dx \\
  2dt + \dfrac{2}{{t - 2}}dt = dx \\
\]
Integrating on both sides, we get
\[
  \int {2dt + \int {\dfrac{{dt}}{{t - 2}} = \int {dx} } } \\
  2t + \ln \left| {t - 2} \right| = x + c{\text{ }}\left[ {\because \int {\dfrac{1}{x} = \ln \left| x \right| + c} } \right] \\
\]
Substituting back \[t = x + y\], We get
\[\therefore 2\left( {x + y} \right) + \ln \left| {x + y - 2} \right| = x + c\]
Thus, the solution of the equation \[\left( {x + y - 1} \right)dx + \left( {2x + 2y - 3} \right)dy = 0\] is \[2\left( {x + y} \right) + \ln \left| {x + y - 2} \right| = x + c\].

Note: A constant namely integrating constant that is added to the function obtained by evaluating the indefinite integral of a given function, indicating that all indefinite integrals of the given function differ by, at most, a constant. So, it is necessary to add integrating constants after completion of the integral.