Question

By eliminating the arbitrary constants A and B from $y = A{x^2} + Bx$ , we get the differential equation:
$A.\dfrac{{{d^3}y}}{{d{x^3}}} = 0 \\\ B.{x^2}\dfrac{{{d^2}y}}{{d{x^2}}} - 2x\dfrac{{dy}}{{dx}} + 2y = 0 \\\$
$C.\dfrac{{{d^2}y}}{{d{x^2}}} = 0$
$D.{x^2}\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0$

Answer 1

Hint: We will solve this question by differentiating the equation with respect to $x$ and keep differentiating, until we get the values of constants A and B. When we get the values we will make some substitutions and this we give us the result.

Complete step-by-step solution -
Here we have,
$y = A{x^2} + Bx.........(1)$

Now, differentiating equation (1) with respect to $x$, we get
$\dfrac{{dy}}{{dx}} = 2Ax + B$ $.........(2)$

Again, differentiating equation (2) with respect to $x$, we get

$$\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( {2Ax + B} \right) \\\ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2A + 0 \\\ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2A \\\ \Rightarrow A = \dfrac{{\dfrac{{{d^2}y}}{{d{x^2}}}}}{2} \\\$$

Now, substituting the value of A in equation (2), we obtain
$\dfrac{{dy}}{{dx}} = 2 \cdot \dfrac{{\dfrac{{{d^2}y}}{{d{x^2}}}}}{2} \cdot x + B \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{d^2}y}}{{d{x^2}}}x + B \\\ \Rightarrow B = \dfrac{{dy}}{{dx}} - \dfrac{{{d^2}y}}{{d{x^2}}}x \\\$

Now, substituting the values of both A and B in equation (1), we obtain
$y = \dfrac{{\dfrac{{{d^2}y}}{{d{x^2}}}}}{2} \cdot {x^2} + \left( {\dfrac{{dy}}{{dx}} - \dfrac{{{d^2}y}}{{d{x^2}}}x} \right)x \\\ \Rightarrow 2y = \dfrac{{{d^2}y}}{{d{x^2}}}{x^2} + 2\left( {\dfrac{{dy}}{{dx}} - \dfrac{{{d^2}y}}{{d{x^2}}}x} \right)x \\\ \Rightarrow 2y = \dfrac{{{d^2}y}}{{d{x^2}}}{x^2} + 2x\dfrac{{dy}}{{dx}} - 2\dfrac{{{d^2}y}}{{d{x^2}}}{x^2} \\\ \Rightarrow 2y = - \dfrac{{{d^2}y}}{{d{x^2}}}{x^2} + 2x\dfrac{{dy}}{{dx}} \\\ \Rightarrow {x^2}\dfrac{{{d^2}y}}{{d{x^2}}} - 2x\dfrac{{dy}}{{dx}} + 2y = 0 \\\$
Hence, option C is the correct answer.

Note: A differential equation is an equation that relates one or more functions and their derivatives. Generally, it defines a relationship between the physical quantities and their rates. These questions must be solved with full concentration as the second derivatives might be confusing.