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

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(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)

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

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(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

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