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

$y = e^{x} (a c o s x + b s i n x) : \frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0$

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Solution

Given, $y = e^{x} (a c o s x + b s i n x)$
On dividing by $e^{x}$ both sides, we get
$e^{- x} y = a c o s x + b s i n x)$ ...(i)
On differentiating both sides w.r.t. x, we get
$e^{- x} \frac{d y}{d x} - y e^{- x} = - a s i n x + b c o s x$
Again, differentiating both sides with respect to x, we get
$e^{- x} \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} e^{- x} - {- y e^{- x} + e^{- x} \frac{d y}{d x}} = - a c o s x - b s i n x \Rightarrow e^{- x} \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} e^{- x} + y e^{- x} - e^{- x} \frac{d y}{d x} = - (a c o s x + b s i n x) \Rightarrow e^{- x} \frac{d^{2} y}{d x^{2}} - 2 e^{- x} \frac{d y}{d x} + y e^{- x} = - y e^{- x}$ [From Eq. (i)]
$\Rightarrow e^{- x} \frac{d^{2} y}{d x^{2}} - 2 e^{- x} \frac{d y}{d x} + 2 y e^{- x} = 0 \Rightarrow e^{- x} {\frac{d^{2} y}{d x 62} - 2 \frac{d y}{d x} + 2 y} = 0 \Rightarrow \frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0$
Hence, the given function is a solution of the corresponding differential equation.

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