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

Hint: 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.