Question

How do you find the general solution to $\dfrac{{dy}}{{dx}} = 2y - 1$?

Answer 1

Hint: Separate same variables on one side and others on the other side. Like put $y$ terms on the left side and then divide both the sides by $dx$. Integrate both the sides. Use of the basic formula of integration which is $\int {\dfrac{1}{x}dx = \log x + c} $ where $c$ is constant.

Complete step-by-step answer:
We are given with the equation $\dfrac{{dy}}{{dx}} = 2y - 1$
Separate the like terms on one side that is move $2y - 1$ from right to left side and we get:
\[\dfrac{{dy}}{{(2y - 1)dx}} = 1\]
Now Multiply both sides by $dx$ so that we are left with only like terms on each side and we get:
\[
  \dfrac{{dy}}{{(2y - 1)dx}}dx = 1 \times dx \\
  \dfrac{{dy}}{{(2y - 1)}} = dx \;
 \]
Since, we have same terms on each side now it becomes easy to solve:
Integrate both the sides and we get:
\[\int {\dfrac{{dy}}{{(2y - 1)}}} = \int {dx} \]
From the formulas of integration we know that $\int {\dfrac{1}{x}dx = \log x + c} $, \[\int {dx} = x + c\], where c is constant:
By using this formula in the equation we get:
\[
  \int {\dfrac{{dy}}{{(2y - 1)}}} = \int {dx} \\
  \dfrac{{\log (2y - 1)}}{2} = x + C \;
 \]
We have taken \[2\] in the denominator because we had \[2\]is the coefficient of $y$and whenever there is a coefficient in $y$ it comes in the denominator in integration.
Now, on further solving the equations;
Multiplying both sides by \[2\] and we get:
\[\log (2y - 1) = 2x + C\], where constant will always remain constant.
Moving \[\log \]from left to right side we get:
\[2y - 1 = {e^{2x + C}}\]
Add both sides by \[1\] and we get:
\[
  2y - 1 + 1 = {e^{2x + C}} + 1 \\
  2y = {e^{2x + C}} + 1 \;
 \]
Dividing both sides by \[2\],we get:\[2\]
\[y = \dfrac{{{e^{2x + C}} + 1}}{2}\].
Therefore, The General Solution of $\dfrac{{dy}}{{dx}} = 2y - 1$ is \[y = \dfrac{{{e^{2x + C}} + 1}}{2}\].
So, the correct answer is “\[y = \dfrac{{{e^{2x + C}} + 1}}{2}\]”.

Note: There are alternate methods also in solving these kinds of equations.
There can be an error if terms are not generated and the integration part is started.
Since this is a general equation, so constant term is necessary.