Solveeit Logo
Login

Question

Question: How do you find the general solution to \(\dfrac{{dy}}{{dx}} = 2y - 1\)?...

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

Explanation

Solution

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

Complete step-by-step answer :
We are given with the equation dydx=2y−1\dfrac{{dy}}{{dx}} = 2y - 1
Separate the like terms on one side that is move 2y−12y - 1 from right to left side and we get:
dy(2y−1)dx=1\dfrac{{dy}}{{(2y - 1)dx}} = 1
Now Multiply both sides by dxdx so that we are left with only like terms on each side and we get:

dy(2y−1)dxdx=1×dx dy(2y−1)=dx   \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:
∫dy(2y−1)=∫dx\int {\dfrac{{dy}}{{(2y - 1)}}} = \int {dx}
From the formulas of integration we know that ∫1xdx=log⁡x+c\int {\dfrac{1}{x}dx = \log x + c} , ∫dx=x+c\int {dx} = x + c, where c is constant:
By using this formula in the equation we get:

∫dy(2y−1)=∫dx log⁡(2y−1)2=x+C   \int {\dfrac{{dy}}{{(2y - 1)}}} = \int {dx} \\\ \dfrac{{\log (2y - 1)}}{2} = x + C \;

We have taken 22 in the denominator because we had 22is the coefficient of yyand whenever there is a coefficient in yy it comes in the denominator in integration.
Now, on further solving the equations;
Multiplying both sides by 22 and we get:
log⁡(2y−1)=2x+C\log (2y - 1) = 2x + C, where constant will always remain constant.
Moving log⁡\log from left to right side we get:
2y−1=e2x+C2y - 1 = {e^{2x + C}}
Add both sides by 11 and we get:

2y−1+1=e2x+C+1 2y=e2x+C+1   2y - 1 + 1 = {e^{2x + C}} + 1 \\\ 2y = {e^{2x + C}} + 1 \;

Dividing both sides by 22,we get:22
y=e2x+C+12y = \dfrac{{{e^{2x + C}} + 1}}{2}.
Therefore, The General Solution of dydx=2y−1\dfrac{{dy}}{{dx}} = 2y - 1 is y=e2x+C+12y = \dfrac{{{e^{2x + C}} + 1}}{2}.
So, the correct answer is “y=e2x+C+12y = \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.