Question

How do I solve the equation \[\dfrac{{dy}}{{dt}} = 2y - 10?\]

Answer 1

Hint: The given question describes the operation of addition/ subtraction/ multiplication/ division. Also, we need to know the process of integration. We need to know how to integrate constant terms and \[\dfrac{1}{x}\] terms. We need to know how to eliminate natural logarithms with exponent components. We would find the value of \[y\] from the given equation.

Complete step-by-step answer:
The given question is shown below,
 \[\dfrac{{dy}}{{dt}} = 2y - 10\]
The above equation can also be written as,
 \[\dfrac{{dy}}{{2y - 10}} = dt \to \left( 1 \right)\]
We would find the value \[y\] from the above equation. For finding the value of \[y\] , let’s integrate the equation \[\left( 1 \right)\] , so we get
 \[\int {\dfrac{{dy}}{{2y - 10}}} = \int {dt} \]
The above equation can also be written as,
 \[\int {\dfrac{{dy}}{{2\left( {y - 5} \right)}}} = \int {dt} \to \left( 2 \right)\]
Let’s solve the LHS part of the above equation,
We get
 \[\int {\dfrac{{dy}}{{2\left( {y - 5} \right)}}} = ?\]
Here \[\dfrac{1}{2}\] is a constant term, so we can take out the integral function. So, we get
 \[\int {\dfrac{{dy}}{{2\left( {y - 5} \right)}}} = \dfrac{1}{2}\int {\dfrac{{dy}}{{y - 5}}} \]
We know that,
 \[\int {\dfrac{1}{y}} dy = \ln y\]
So, we get
 \[\dfrac{1}{2}\int {\dfrac{{dy}}{{y - 5}}} = \dfrac{1}{2}\ln \left( {y - 5} \right) \to \left( 3 \right)\]
Let’s solve the RHS part of the equation \[\left( 2 \right)\] , we have,
 \[\int {dt} = ?\]
We know that,
 \[\int {dy} = y + C\]
So, we get
 \[\int {dt} = t + C \to \left( 4 \right)\]
By substituting the equation \[\left( 3 \right)\] and \[\left( 4 \right)\] in the equation \[\left( 2 \right)\] , we get
 \[
  \left( 2 \right) \to \int {\dfrac{{dy}}{{2\left( {y - 5} \right)}}} = \int {dt} \\
  \dfrac{1}{2}\ln \left( {y - 5} \right) = t + C \;
 \]
The above equation can also be written as,
 \[
  \ln \left( {y - 5} \right) = 2(t + C) \\
  \ln (y - 5) = 2t + 2C \;
 \]
The above equation can be modified as follows,
 \[\ln (y - 5) = 2t + C\]
To solve the above equation, let’s take the exponent on both sides of the above equation,
 \[{e^{\ln \left( {y - 5} \right)}} = {e^{2t}} + {e^c} \to \left( 5 \right)\]
Let’s solve the above equation,
 \[{e^{\ln \left( {y - 5} \right)}} = ?\]
We know that exponent function can be canceled with logarithmic function so we get,
 \[{e^{\ln \left( {y - 5} \right)}} = \left( {y - 5} \right)\]
We know that,
So, we get
Here, \[{e^c} = A\]
So, we get
 \[{e^{2t + c}} = A{e^{2t}}\]
So, the equation \[\left( 5 \right)\] becomes,
 \[
  \left( 5 \right) \to {e^{\ln \left( {y - 5} \right)}} = {e^{2t}} + {e^c} \\
  y - 5 = A{e^{2t}} \\
  y = 5 + A{e^{2t}} \to \left( 6 \right) \;
 \]
So, the final answer is,
 \[y = 5 + A{e^{2t}}\]
So, the correct answer is “ \[y = 5 + A{e^{2t}}\] ”.

Note: Remember the basic formulae for the natural algorithm. Note that, when the natural logarithm and exponent are involved in a single term \[x\] we can cancel the \[\ln \] and \[{e^x}\] with each other, and the answer will be \[x\] . Note that when any number can be added with a constant term \[C\] , the term \[C\] won’t change it remains as \[C\] . This type of question involves the operation of addition/ subtraction/ multiplication/ division.