For the question given below verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

$y = a e^{x} + b e^{- x} + x^{2}; x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0$

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Solution

Given, $y = a e^{x} + b e^{- x} + x^{2}; x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0$ ....(i)
and $y = a e^{x} + b e^{- x} + x^{2}$
On differentiating both sides of Eq.(ii) w.r.t. x, we get
$\frac{d y}{d x} = a e^{x} - b e^{- x} + 2 x$
Again, differentiating both sides w.r.t. x, we get
$\frac{d^{2} y}{d x^{2}} = a e^{x} + b e^{- x} + 2$
Now, on substituting the values of y, $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ in Eq. (i), we get
$L H S = x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = x (a e^{x} + b e^{- x} + 2) + 2 (a e^{x} - b e^{- x} + 2 x) - x (a e^{x} + b e^{- x} + x^{2}) + x^{2} - 2 = (a x e^{x} + b x e^{- x} + 2 x) + (2 a e^{x} - 2 b e^{- x} + 4 x) - (a x e^{x} + b x e^{- x} + x^{3}) + x^{2} - 2 = a x e^{x} + b x e^{- x} + 2 x + 2 a e^{x} - 2 b e^{- x} + 4 x - a x e^{x} - b x e^{- x} - x^{3} + x^{2} - 2 = 2 a e^{x} - 2 b e^{- x} - x^{3} + x^{2} + 6 x - 2 \neq 0 L H S \neq R H S$
Hence, the given function is not a solution of the corresponding differential equation.

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For the question given below verify that the given function (implicit or explicit) is a solution of the corresponding differential equation. y=aex+be−x+x2;xd2ydx2+2dydx−xy+x2−2=0

For the question given below verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

$y = a e^{x} + b e^{- x} + x^{2}; x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0$