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.