The differential equation of the family of curves $y = Ae^{3x} + Be^{5x}$ where A and B are arbitrary constants, is

[MNR 1988]

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Solution

The correct option is B

$y = A e^{3 x} + B e^{5 x} \Rightarrow \frac{d y}{d x} = 3 A e^{3 x} + 5 B e^{5 x} \Rightarrow \frac{d^{2} y}{d x^{2}} = 9 A e^{3 x} + 25 B e^{5 x}$
$\Rightarrow \frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} + 15 y = 0$ (By inspection)

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Similar questions

The differential equation of the family of curves $y = A e^{3 x} + B e^{5 x}$ ,where A and B are arbitrary constants, is

The differential equation of the family of curves where A and B are arbitrary constants, is

[MNR 1988]

y = A e^{3 x} + B e^{5 x}

A

B

y = a e^{- 3 x} + b,

a

b

\frac{d^{2} y}{d x^{2}} + 3 \frac{d y}{d x} - 2 y = 0

y = a cos (x + b)

y^{^{'}} = \frac{d y}{d x}

y^{^{''}} = \frac{d^{2} y}{d x^{2^{^{'}}}}

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