What is a general solution to the differential equation $ y' - 3y = 5 $ ?

Hint: The given equation is a first order differential equation and we have to find the the general solution of this equation. First of all we will write y’ as $ \dfrac{{dy}}{{dx}} $ and then take all the y terms with dy and all the x terms with dx. Then we will integrate on both sides and by simplifying, we will get the general solution.

Complete step by step solution:
In this question, we are supposed to find the general solution of the differential equation $ y' - 3y = 5 $ .
This is a first order differential equation.
First of all, y’ means $ \dfrac{{dy}}{{dx}} $ . So, we can write our equation as
 $ \Rightarrow \dfrac{{dy}}{{dx}} - 3y = 5 $ - - - - - - - - - (1)
Now, we need to separate the terms dy and dx. So for that take 3y to the other side of the equal to sign.
Therefore, equation (1) becomes,
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \left( {5 + 3y} \right) $ - - - - - - - - - - (2)
Now, take the y terms with dy and x terms with dx. But, in equation (2), there are no x terms. So just take 1 with dx and y terms with dy.
Therefore, equation (2) becomes,
 $ \Rightarrow \dfrac{1}{{3y + 5}}dy = 1dx $ - - - - - - - - - - (3)
Integrating on both sides, equation (3) becomes
 $ \Rightarrow \int {\dfrac{1}{{3y + 5}}dy = \int {1dx} } $
Now, we know that $ \int {\dfrac{1}{x} = \ln x} $ and $ \int {1 = } x $
Therefore, above equation becomes
 $ \Rightarrow \dfrac{1}{3}\ln \left( {3y + 5} \right) = x + c $ - - - - - - - - (4)
Where c is the integration constant.
Take 3 to RHS, we get
 $
   \Rightarrow \ln \left( {3y + 5} \right) = 3\left( {x + c} \right) \\
   \Rightarrow 3y + 5 = {e^{3\left( {x + c} \right)}} \\
   \Rightarrow 3y = {e^{3\left( {x + c} \right)}} - 5 \\
   \Rightarrow y = \dfrac{{\left( {{e^{3\left( {x + c} \right)}} - 5} \right)}}{3} \;
  $
Hence, the general solution of the differential equation $ y' - 3y = 5 $ is $ y = \dfrac{{\left( {{e^{3\left( {x + c} \right)}} - 5} \right)}}{3} $ .
So, the correct answer is “ $ y = \dfrac{{\left( {{e^{3\left( {x + c} \right)}} - 5} \right)}}{3} $ .”.

Note: Note that we have found the general equation of the given equation. In order to find the particular solution, we need to find the value of c.
Suppose we are given that the value of x is 2 at $ y = 3 $ . Put these values in the general solution and find the value of c. After finding the value of c put that value in the general solution and you will get the particular solution.