The differential equation of which xy = ae^x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given equation is $xy = ae^x + be^{-x} + x^2$  $(i)$ Differentiating both sides with respect to $x$, we get: $y + xy' = ae^x - be^{-x} + 2x$  (ii)

Vedclass · Accepted Answer

$xy'' + 2y' - xy + x^2 - 2 = 0$

Vedclass · Answer

(D) Given equation is $xy = ae^x + be^{-x} + x^2$  $(i)$ Differentiating both sides with respect to $x$, we get: $y + xy' = ae^x - be^{-x} + 2x$  (ii) Differentiating again with respect to $x$, we get: $y' + y' + xy'' = ae^x + be^{-x} + 2$ $2y' + xy'' = ae^x + be^{-x} + 2$  (iii) From $(i)$, we have $ae^x + be^{-x} = xy - x^2$. Substituting this into (iii): $xy'' + 2y' = (xy - x^2) + 2$ Rearranging the terms, we get: $xy'' + 2y' - xy + x^2 - 2 = 0$

The differential equation of which $xy = ae^x + be^{-x} + x^2$ is a solution, is

Similar Questions

The order of the differential equation of the family of all concentric circles centered at $(h, k)$ is

Find the differential equation for the family of curves given by $y = a e^{3x} + b e^{-2x}$ by eliminating the arbitrary constants $a$ and $b$.

The differential equation of all parabolas whose axes are parallel to the $y$-axis is

Form the differential equation representing the family of curves given by $(x-a)^{2}+2 y^{2}=a^{2},$ where $a$ is an arbitrary constant.

The differential equation of the family of curves represented by the equation ${x^2} + {y^2} = {a^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module