Question

Given, $y\left( 0 \right) = 2000,{\text{ }}\dfrac{{dy}}{{dx}} = 32000 - 20{y^2}$ then what is the value of $\mathop {\lim }\limits_{x \to \infty } y\left( x \right)$.
${\text{A}}{\text{. 20}} \\\ {\text{B}}{\text{. 40}} \\\ {\text{C}}{\text{. 60}} \\\ {\text{D}}{\text{. 320}} \\\$

Answer 1

We will first integrate the given equation in order to find the value of y, then we will find the value of C using the value of y when x=0, then we will get the appropriate response by simplifying the obtained equation.

Complete step-by-step answer :
Given that $y\left( 0 \right) = 2000,{\text{ }}\dfrac{{dy}}{{dx}} = 32000 - 20{y^2}$
By cross multiplying the equation will become as

$$\dfrac{{dy}}{{dx}} = 20\left( {1600 - {y^2}} \right) \\\ \Rightarrow \dfrac{{dy}}{{1600 - {y^2}}} = 20dx \\\ \Rightarrow \dfrac{{dy}}{{ - \left( {{y^2} - 1600} \right)}} = 20dx \\\ \Rightarrow \dfrac{{dy}}{{{y^2} - 1600}} = - 20dx \\\$$

As, ${y^2} - 1600 = \left( {y + 40} \right)\left( {y - 40} \right)$
Now we will add $y + 40$ and subtract $y - 40$ in the numerator of the LHS of above equation, we have

$$\Rightarrow \left[ {\dfrac{1}{{80}}} \right]\left[ {\int {\left( {\dfrac{{y + 40}}{{\left( {y + 40} \right)\left( {y - 40} \right)}} - \dfrac{{y - 40}}{{\left( {y + 40} \right)\left( {y - 40} \right)}}} \right)} dy} \right] = - 20\int {dx} \\\ \Rightarrow \int {\left( {\dfrac{1}{{y - 40}} - \dfrac{1}{{y + 40}}} \right)} dy = - 80 \times 20\int {dx} \\\ \Rightarrow \int {\dfrac{1}{{y - 40}}dy} - \int {\dfrac{1}{{y + 40}}dy} = - 1600\int {dx} \\\$$

Also, we know that $\int {\dfrac{1}{{y - b}}dy} = \log \left( {y - b} \right)$
$\Rightarrow \log \left( {y - 40} \right) - \log \left( {y + 40} \right) = - 1600x + {\text{C}}$ where C represents the constant of integration
Using the identity $\log m - \log n = \log \dfrac{m}{n}$ in the above equation, we get
$\Rightarrow \log \left( {\dfrac{{y - 40}}{{y + 40}}} \right) = - 1600x + {\text{C }} \to {\text{(1)}}$
Now we will find out the value of C by putting the value of y as 2000 when x = 0 because it is given that $y\left( 0 \right) = 2000$

$$\Rightarrow \log \left( {\dfrac{{2000 - 40}}{{2000 + 40}}} \right) = - 20\left( 0 \right) + {\text{C}} \\\ \Rightarrow {\text{C}} = \log \left( {\dfrac{{1960}}{{2040}}} \right) \\\ \Rightarrow {\text{C}} = \log \left( {\dfrac{{49}}{{51}}} \right) \\\$$

By putting the value of constant of integration C obtained in the equation (1), we get

$$\Rightarrow \log \left( {\dfrac{{y - 40}}{{y + 40}}} \right) = - 1600x + \log \left( {\dfrac{{49}}{{51}}} \right) \\\ \Rightarrow \log \left( {\dfrac{{y - 40}}{{y + 40}}} \right) - \log \left( {\dfrac{{49}}{{51}}} \right) = - 1600x \\\$$

Using the identity $\log m - \log n = \log \dfrac{m}{n}$ in the above equation, we get

$$\Rightarrow \log \left( {\dfrac{{y - 40}}{{y + 40}} \times \dfrac{{51}}{{49}}} \right) = - 20x \\\ \Rightarrow - \log \left[ {\dfrac{{51\left( {y - 40} \right)}}{{49\left( {y + 40} \right)}}} \right] = 20x \\\$$

Using the formula $- \log a = \log \left( {\dfrac{1}{a}} \right)$ in the above equation, we get
$\Rightarrow \log \left[ {\dfrac{{49\left( {y + 40} \right)}}{{51\left( {y - 40} \right)}}} \right] = 20x$
As, $x \to \infty \Rightarrow 20x \to \infty$
Now for $x \to \infty$, the RHS part of the above equation will be equal to infinity. To make the LHS part of the equation equal to infinity below condition must be satisfied
$\log \left[ {\dfrac{{49\left( {y + 40} \right)}}{{51\left( {y - 40} \right)}}} \right] \to \infty$
Using the formula $\log y \to \infty \Rightarrow y \to \infty$, we get
$\dfrac{{49\left( {y + 40} \right)}}{{51\left( {y - 40} \right)}} \to \infty$
The term in the LHS of the above will approach infinity only when the denominator approaches to zero because whenever any number is divided by zero it approaches infinity
$51\left( {y - 40} \right) \to 0 \\\ y - 40 \to 0 \\\ y \to 40 \\\$
Therefore, the required value of y is y = 40.
Hence, the correct answer is option B.

Note : A differential equation, in mathematics, is an equation that relates one or more functions and their derivatives. The functions typically represent physical quantities in applications, the derivatives represent their rate of change and the differential equation describes a relation between the two.