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

Hint: 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.