Question

The solution of the differential equation
$x = 1 + xy\dfrac{{dy}}{{dx}} + \dfrac{{{x^2}{y^2}}}{{2!}}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} + \dfrac{{{x^3}{y^3}}}{{3!}}{\left( {\dfrac{{dy}}{{dx}}} \right)^3} + \cdot \cdot \cdot \cdot \cdot \infty $ is
A. ${y^2} = lnx + c$
B. $y = {\left( {\ln x} \right)^2} + c$
C. $y = \pm \ln x + c$
D. ${y^2} = {\left( {\ln x} \right)^2} + 2c$

Answer 1

Hint: As we know that the given differential equation is an infinite series so first recognize the type of series and then solve it with the help of the formulas of integration. Use the logarithm concept to get the correct answer.

Complete step-by-step solution:
The given question is based on a differential equation. The main purpose of the differential equation is to find the function over its entire domain. It is used to describe the exponential growth or decay over time. It has the probability to predict the world in our surroundings. It is widely used in various fields such as, physics, chemistry, biology, economics, and so on.
There are two main types of differential equation – Partial differential equation and Ordinary differential equation.
The given differential equation is written as,
$x = 1 + xy\dfrac{{dy}}{{dx}} + \dfrac{{{x^2}{y^2}}}{{2!}}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} + \dfrac{{{x^3}{y^3}}}{{3!}}{\left( {\dfrac{{dy}}{{dx}}} \right)^3} + \cdot \cdot \cdot \cdot \cdot \infty $
The given differential is the type of the infinite series and it is the type of exponential series as shown below.
$
  {e^x} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \cdot \cdot \cdot \cdot \cdot \\
    \\
 $
So,
The given differential equation can be written as,
$x = {e^{xy \cdot \dfrac{{dy}}{{dx}}}}......\left( 1 \right)$
Now, Taking log on both sides in equation (1).
$\ln x = x \cdot y\dfrac{{dy}}{{dx}}$
Now on integration the above equation and it can be written as,
$\smallint \dfrac{{\ln x}}{x}dx + c = \smallint ydy$
Let, $\ln x = t,\dfrac{{dx}}{x} = dt$
So, it is calculated as,
$\smallint t \cdot dt + c = \dfrac{{{y^2}}}{2}$
After integration we get,
$ \Rightarrow \dfrac{{{t^2}}}{2} + c = \dfrac{{{y^2}}}{2}$
Now we substitute $\ln x$ for $t$ in the above expression.
${\dfrac{{\left( {\ln x} \right)}}{2}^2} + c = \dfrac{{{y^2}}}{2}$
After simplification we obtain,
$\therefore {\left( {\ln x} \right)^2} + 2c = {y^2}$
Therefore, it is the required solution of the given differential equation.

Hence, the correct option is D.

Note: As we know that a differential equation represents a relationship between one or more functions and their derivatives. In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the functions and their derivatives.