Question

The D.E. whose solution is $xy = a{x^2} + \dfrac{b}{x}$ is
A.${x^2}{y_2} + 2x{y_1} = 2y$
B.${x^2}{y_2} - x{y_1} + 2y = 0$
C.${x^2}{y_2} + x{y_1} + y = 0$
D.${x^2}{y_2} + x{y_1} + 2y = 0$

Answer 1

Here, we will first differentiate the given solution with respect to $x$ to find the first value of $a$. We will again differentiate it with respect to $x$ to find the second value of $a$. Then we will equate both the values of $a$ and solve it further to find the required differential equation.

Formula Used:
We will use the following formulas:
1.$\dfrac{{dy}}{{dx}}\left( {A \cdot B} \right) = A \cdot \left( {\dfrac{{dy}}{{dx}}B} \right) + B \cdot \left( {\dfrac{{dy}}{{dx}}A} \right)$
2.$\dfrac{{dy}}{{dx}}{x^n} = n{x^{n - 1}}$

Complete step-by-step answer:
We are given that the solution of a differential equation is $xy = a{x^2} + \dfrac{b}{x}$
First, we will take the LCM in the RHS. Therefore, we get
$\Rightarrow xy = \dfrac{{a{x^3} + b}}{x}$
Multiplying both sides by $x$, we get
$\Rightarrow {x^2}y = a{x^3} + b$
Now, differentiating both sides with respect to $x$ using the formula of differentiation $\dfrac{{dy}}{{dx}}\left( {A \cdot B} \right) = A \cdot \left( {\dfrac{{dy}}{{dx}}B} \right) + B \cdot \left( {\dfrac{{dy}}{{dx}}A} \right)$ and $\dfrac{{dy}}{{dx}}{x^n} = n{x^{n - 1}}$, we get
$\Rightarrow {x^2}\left( {\dfrac{{dy}}{{dx}}} \right) + y\left( {2x} \right) = a\left( {3{x^2}} \right)$
Multiplying the terms, we get
$\Rightarrow {x^2}\left( {\dfrac{{dy}}{{dx}}} \right) + 2xy = 3a{x^2}$
Here, substituting $\dfrac{{dy}}{{dx}} = {y_1}$ in the above equation, we get
$\Rightarrow {x^2}{y_1} + 2xy = 3a{x^2}$
Dividing both sides by $3{x^2}$, we get
$\Rightarrow \dfrac{{{x^2}{y_1} + 2xy}}{{3{x^2}}} = a$
Taking $x$common from the numerator and cancelling it out with that of the denominator, we get,
$\Rightarrow a = \dfrac{{x{y_1} + 2y}}{{3x}}$………………………………….$\left( 1 \right)$
Multiplying both sides by $3x$, we get
$\Rightarrow 3xa = x{y_1} + 2y$
Now back substituting $\dfrac{{dy}}{{dx}} = {y_1}$ in the above equation, we get
$\Rightarrow 3xa = x\left( {\dfrac{{dy}}{{dx}}} \right) + 2y$
Hence, again, differentiating both sides with respect to $x$, we get
$\Rightarrow 3a\left( 1 \right) = x\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) + \left( {\dfrac{{dy}}{{dx}}} \right)\left( 1 \right) + 2\left( {\dfrac{{dy}}{{dx}}} \right)$
Again, writing $\dfrac{{dy}}{{dx}} = {y_1}$ and $\dfrac{{{d^2}y}}{{d{x^2}}} = {y_2}$, we get
$\Rightarrow 3a = x{y_2} + {y_1} + 2{y_1}$
Adding the like terms, we get
$\Rightarrow 3a = x{y_2} + 3{y_1}$
Dividing both sides by 3, we get
$\Rightarrow a = \dfrac{{x{y_2} + 3{y_1}}}{3}$…………………………….$\left( 2 \right)$
Now, we have two values of $a$ so will equate equation $\left( 1 \right)$ and $\left( 2 \right)$. Therefore, we get
$\dfrac{{x{y_1} + 2y}}{{3x}} = \dfrac{{x{y_2} + 3{y_1}}}{3}$
Multiplying both sides by $3x$, we get
$\Rightarrow x{y_1} + 2y = x\left( {x{y_2} + 3{y_1}} \right)$
$\Rightarrow x{y_1} + 2y = {x^2}{y_2} + 3x{y_1}$
Solving further, we get
$\Rightarrow 2y = {x^2}{y_2} + 3x{y_1} - x{y_1}$
Subtracting the like terms, we get
$\Rightarrow {x^2}{y_2} + 2x{y_1} = 2y$
Therefore, the D.E. whose solution is $xy = a{x^2} + \dfrac{b}{x}$ is ${x^2}{y_2} + 2x{y_1} = 2y$
Hence, option A is the correct answer.

Note: In mathematics, differential equations are those equations which relate to one or more than one functions and their respective derivatives. The differential equation defines a relationship between the functions which are representing a physical quantity and the derivatives which represents the rate of change of those physical quantities. Also, the first derivative represents the slope of the function at a point graphically, whereas, the second order derivative shows how the slope changes over the independent variable in the graph.