From the given equation, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

$y = e^{x} (a c o s x + b s i n x)$

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Solution

Given, $y = e^{x} (a c o s x + b s i n x)$ ...(i)
On dividing both sides by $e^{x}, e^{- x} y = (a c o s x + b s i n x)$ ...(ii)
On differentiating both sides w.r.t. x, we get
$e^{- x} y^{'} + y e^{- x} (- 1) = - a s i n x + b c o s x$
[Using product rule , $\frac{d}{d x} (u . v) = u \frac{d}{d x} v + v \frac{d}{d x} u$ ]
Again differentaiting both sides w.r.t. x, we get
$e^{- x} \frac{d}{d x} (y^{'}) + y^{'} \frac{d}{d x} (e^{- x}) - [y \frac{d}{d x} e^{- x} + e^{- x} \frac{d}{d x} y] = - a c o s x - b s i n x \Rightarrow e^{- x} y^{''} - 2 y^{'} e^{- x} + y e^{- x} = - (y e^{- x}) [u s i n g E q . (i i)] \Rightarrow e^{- x} [y^{''} - 2 y^{'} + 2 y] = 0 \Rightarrow y^{''} - 2 y^{'} + 2 y = 0 (d i v i d i n g b y e^{- x})$
which is the required differential equation.

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