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.