Question

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

Answer 1

Here, we will first differentiate the given equation w.r.t $x$ and then differentiate again w.r.t $x$ to find the second order differential equation. Then we will solve the equations by illuminating the constants and in the end we’ll take ${y_2} = \dfrac{{{d^2}y}}{{d{x^2}}}$ and ${y_1} = \dfrac{{dy}}{{dx}}$ in the obtained equation for the required differential equation.

Complete step by step solution: We are given that the solution is $xy = a{e^x} + b{e^{ - x}} + {x^2}$.

First, we will differentiate the given equation with respect to $x$.

$\Rightarrow x\dfrac{{dy}}{{dx}} + y = a{e^x} - b{e^{ - x}} + 2x$

Differentiating the above equation again with respect to $x$, we get

$\Rightarrow \dfrac{{dy}}{{dx}} + x\dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{{dy}}{{dx}} = a{e^x} + b{e^{ - x}} + 2$

Combining the like terms in the equation, we get

$\Rightarrow x\dfrac{{{d^2}y}}{{d{x^2}}} + 2\dfrac{{dy}}{{dx}} = xy - {x^2} + 2$

We will now take ${y_2} = \dfrac{{{d^2}y}}{{d{x^2}}}$ and ${y_1} = \dfrac{{dy}}{{dx}}$ in the above equation, we get

$x{y_2} + 2{y_1} = xy + 2 - {x^2}$

Thus, the required differential equation is $x{y_2} + 2{y_1} = xy + 2 - {x^2}$.

Hence, option B is correct.

Note: Whenever we need to find the differential equation of a given equation, we always focus on illuminating the constants by differentiating the equation. In this question we had 2 constants so to illuminate 2 constants we need 2 equation and hence we differentiated it two times.