The value of $m_{1} + m_{2} + m_{3}$ & $m_{1}$ , $m_{2}$ , $m_{3}$ are respectively

9, 15

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15, 9

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- 9, - 15

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None of these

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Solution

The correct option is A $9, 15$
As $m_{1}$ , $m_{2}$ , $m_{3}$ are roots of $m^{3} - 9 m^{2} + 23 m - 15 = 0$
$\Rightarrow$ $(m - 1) (m - 3) (m - 5) = 0$
$\Rightarrow$ $m_{1} = 1$ , $m_{2} = 3$ , $m_{3} = 5$
$∴$ Required solution is $y = c_{1} e^{x} + c_{2} e^{3 x} + c_{3} e^{5 x}$ (i)
On Differentiating w.r to x both sides we get
$\Rightarrow$ $y^{'} = c_{1} e^{x} + 3 c_{2} e^{3 x} + 5 c_{3} e^{5 x}$ (ii)
(using (ii) -(i))
$\Rightarrow$ $y^{'} - y = 2 c_{2} e^{3 x} + 4 c_{3} e^{5 x}$ (iii)
Again differentiating w.r to x both sides
$\Rightarrow$ $y^{''} - y^{'} = 6 c_{2} e^{3 x} + 20 c_{3} e^{5 x}$ (iv)
using (iv)-3(iii) we get
$y^{''} - 4 y^{'} + 3 y = 8 c_{3} e^{5 x}$ (v)
$\Rightarrow$ $y^{'''} - 4 y^{''} + 3 y^{'} = 8 \times 5 c_{3} e^{5 x}$ (vi)
using (vi)-5(v) we get
$y^{'''} - 9 y^{''} + 23 y^{'} - 15 y = 0$
or $\frac{d^{3} y}{d x^{3}} - 9 \frac{d^{2} y}{d x^{2}} + 23 \frac{d y}{d x} - 15 y = 0$
$= A \frac{d^{3} y}{d x^{3}} + B \frac{d^{2} y}{d x^{2}} + C \frac{d y}{d x} + D$
$\Rightarrow$ A=1, B=-9, C=23. D=-15
We obtained the following differential equation
$\frac{d^{3} y}{d x^{3}} - 9 \frac{d^{2} y}{d x^{2}} + 23 \frac{d y}{d x} - 15 y = 0$
$m_{1}$ , $m_{2}$ , $m_{3}$ are roots of the equation
$m^{3} - 9 m^{2} + 23 m - 15 = 0$
$∴$ $m_{1} + m_{2} + m_{3} = 9$ & $m_{1} m_{2} m_{3} = 15$
$∴$ $(m_{1} + m_{2} + m_{3}, m_{1} m_{2} m_{3}) = (9, 15)$

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