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

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Solution

$\Rightarrow y = a e^{- 27 + b e^{3 x}} - - - - (i)$
$\Rightarrow y_{1} = - 2 a e^{- 2 x} + 3 b e^{3 x} - - - - (i i)$
$\Rightarrow y_{2} = 4 a e^{- 27} + 9 b e^{3 x} - - - - - (i i i)$

$E q^{n} (i i) + 2 x E q^{n} (i) w e g e t$ $E q^{n} (i i i) - 3 E q^{n} (i i) w e g e t$
$y_{1} + 2_{y} = 5 b e^{3 x}$ $y_{2} - 3_{y} = 10 a e^{- 2 x}$
$\Rightarrow [b e^{3 x} = \frac{1}{5} (y_{1} + 2 y_{y})]$ $[a e^{- 2 x} = \frac{1}{10} (y_{2} - 3_{y 1})]$
putting vasues of $a e^{- 2 x}$ & $b e^{3 x} i n e q^{n} (i)$
$y = \frac{1}{5} (y_{1} + 2_{y}) + \frac{1}{10} (y_{2} - 3_{y 1})$
$\Rightarrow 10_{y} = 2_{y 1} + 4_{y} + y_{2} - 3_{y 1}$
$\Rightarrow 10_{y} = 5_{y 2} - y_{1}$
$\Rightarrow [10_{y} + y_{1} - 5_{y 2} = 0]$

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