The D.E whose solution is $y = a x + b e^{x}$ :

(x - 1) y_{2} - x y_{1} + y = 0

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(x - 1) y_{2} + x y_{1} = y

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x^{2} y_{2} - x y_{1} + y = 0

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x^{2} y_{2} + x y_{1} - y = 0

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Solution

The correct option is A $(x - 1) y_{2} - x y_{1} + y = 0$
$y = a x + b e^{x}$
$\Rightarrow y_{1} = a + b e^{x}$
$\Rightarrow y_{2} = b e^{x}$
$\Rightarrow y_{1} = a + y_{2}$
$a = (y_{1} - y_{2})$
$\Rightarrow y = (y_{1} - y_{2}) x + y_{2}$
$\to y_{2} (x - 1) - y_{1} x + y = 0$

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Similar questions

If y = 3 cos (log x)+4 sin (log x), show that $x^{2} y_{2} + x y_{1} + y = 0.$

x = \sin (\frac{1}{a} \log y)

y = {sin}^{- 1} x

(1 - x^{2}) y_{2} - x y_{1} = 0

x = sin t

y = sin p t,

(1 - x^{2}) y_{2} - x y_{1} + p^{2} y = 0

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