Obtain the differential equation by eliminating the arbitrary constants from the following equations:
$y = A e^{3 x} + B . e^{- 3 x}$

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Solution

Given,
$y = A e^{3 x} + B e^{- 3 x}$
$\frac{d y}{d x} = 3 A e^{3 x} - 3 B e^{- 3 x}$
$\frac{d^{2} y}{d x^{2}} = 9 (A e^{3 x} + B e^{- 3 x})$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = 9 (A e^{3 x} + B e^{- 3 x})$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = 9 y$
$∴ \frac{d^{2} y}{d x^{2}} - 9 y = 0$

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Obtain the differential equation by eliminating the arbitrary constants from the following equations:y=Ae3x+B.e−3x

Given, y=Ae3x+Be−3xdydx=3Ae3x−3Be−3xd2ydx2=9(Ae3x+Be−3x)⇒d2ydx2=9(Ae3x+Be−3x)⇒d2ydx2=9y∴d2ydx2−9y=0

Obtain the differential equation by eliminating the arbitrary constants from the following equations:
$y = A e^{3 x} + B . e^{- 3 x}$

Given,
$y = A e^{3 x} + B e^{- 3 x}$
$\frac{d y}{d x} = 3 A e^{3 x} - 3 B e^{- 3 x}$
$\frac{d^{2} y}{d x^{2}} = 9 (A e^{3 x} + B e^{- 3 x})$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = 9 (A e^{3 x} + B e^{- 3 x})$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = 9 y$
$∴ \frac{d^{2} y}{d x^{2}} - 9 y = 0$