Form a differential equation representing the given family
of curves by eliminating arbitrary constant $a$ and $b$ .
$y = a e^{3 x} + b e^{- 2 x}$

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Solution

$y = a e^{3 x} + b e^{- 2 x}$
Differentiating we get,
$y^{^{'}} = 3 a e^{3 x} - 2 b e^{- 2 x}$ .......(i)
Differentiating we get,
$y^{^{''}} = 9 a e^{3 x} + 4 b e^{- 2 x}$
Subtracting (i)
$y^{^{''}} - y^{^{'}} = 9 a e^{3 x} + 4 b e^{^{'}} - 3 a e^{3 x} + 2 b e^{- 2 x} = 6 (a e^{3 x} + b e^{- 2 x})$
$y^{^{''}} - y^{^{'}} = 6 y$

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Q. In each of the Exercises

1

5,

form a differential equation representing the given family of curves by eliminating arbitrary constants

a

and

b .

(1)

\frac{x}{a} + \frac{y}{b} = 1

(2)

y^{2} = a (b^{2} - x^{2})

(3)

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Form a differential equation representing the given familyof curves by eliminating arbitrary constant a and b.y=ae3x+be−2x

y=ae3x+be−2xDifferentiating we get,y′=3ae3x−2be−2x.......(i)Differentiating we get,y′′=9ae3x+4be−2xSubtracting (i)y′′−y′=9ae3x+4be′−3ae3x+2be−2x=6(ae3x+be−2x)y′′−y′=6y

Form a differential equation representing the given family
of curves by eliminating arbitrary constant $a$ and $b$ .
$y = a e^{3 x} + b e^{- 2 x}$