Approach matters here, show yours

Given Primitive : $y=ae^{x}+be^{2x}+ce^{3x}$

$\implies y.e^{-x}=a+be^x+ce^{2x}$

Differentiating wrt x

$e^{-x}\left(\frac{dy}{dx}-y\right)=be^{x}+2ce^{2x}$

$\implies e^{-2x}\left(\frac{dy}{dx}-y\right)=b+2ce^{x}$

Differentiating wrt x again

$e^{-2x}\left(\frac{d^{2}y}{dx^{2}}-3.\frac{dy}{dx}+2y\right)=2ce^{x}$

$\implies e^{-3x}\left(\frac{d^{2}y}{dx^{2}}-3.\frac{dy}{dx}+2y\right)=2c$

Differentiating wrt x again, we have

$e^{-3x}\left(\frac{d^{3}y}{dx^{3}}-6.\frac{d^{2}y}{dx^{2}}+11.\frac{dy}{dx}-6y\right)=0$

As $e^{-3x} \neq 0$

Therefore Required differential equation is

$\frac{d^{3}y}{dx^{3}}-6.\frac{d^{2}y}{dx^{2}}+11.\frac{dy}{dx}-6y=0$

On comparing coefficients we get $\dfrac{b \times c}{a \times d}=\boxed{11}$ .

$y=ae^{x}+be^{2x}+ce^{3x}$

$\frac{dy}{dx}=ae^{x}+2be^{2x}+3ce^{3x}$

$\frac{d^{2}y}{dx^{2}}=ae^{x}+4be^{2x}+9ce^{3x}$

$\frac{d^{3}y}{dx^{3}}=ae^{x}+8be^{2x}+27ce^{3x}$

Substituting in given equation,

$a\frac{dy}{dx}+b\frac{d^{2}y}{dx^{2}}+c\frac{dy}{dx}+dy=0$

$\Rightarrow a(a+b+c+d)e^{x}+b(8a+4b+2c+d)e^{2x}+c(27a+9b+3c+d)e^{3x}=0$

Since $e^{x}$ , $e^{2x}$ and $e^{3x}$ cannot be 0 or negative, their coefficients must be 0.

$a+b+c+d=0$ ..... $\color{#D61F06}{(1)}$

$8a+4b+2c+d=0$ ..... $\color{#D61F06}{(2)}$

$27a+9b+3c+d=0$ ..... $\color{#D61F06}{(3)}$

$\color{#D61F06}{(2)-(1)}$ $7a+3b+c=0$ ..... $\color{#D61F06}{(4)}$

$\color{#D61F06}{(3)-(2)}$ $19a+5b+c=0$ ..... $\color{#D61F06}{(5)}$

$\color{#D61F06}{(5)-(4)}$ $12a+2b=0$ ..... $\color{#D61F06}{(6)}$

Let $a=1$ . It gives $b=-6,\ c=11,\ d=-6$

Hence $\frac{b×c}{a×d}=11$

$\color{#3D99F6}{\text{There is no need to substitute a=1. You may get all other variables in terms of a. While calculating fraction, a would cancel out.}}$

we know from superposition that each of $e^{x}$ $e^{2x}$ and $e^{3x}$ are solutions to the differential equation. Hence, 1,2 and 3{powers of solutions} are solutions to the characteristic eq. of the differential equation
which is.............. a $r^{3}$ + b $r^{2}$ + cr + d = 0 ( cubic equation ) we already know the roots (1,2, and 3) and by using vietas' formulas we find the answer to be of the form .....

$b \times c$ / $a \times d$ = [{sum of roots} x {sum taken two at a time}]/{product of roots} which when computed gives 11 as the answer.

i sued determinants without wasting time finding a,b,c,d separately but a c/b d directly

Done exactly the same as Sandeep

$y = ae^x + be^{2x} + ce^{3x} \color{#20A900}{. . . . . . . (1)}$

$y' = ae^x + 2be^{2x} + 3ce^{3x} = (ae^x + be^{2x} + ce^{3x}) + be^{2x} + 2ce^{3x}$

Using $\color{#20A900}{(1)}$

$y' = y + be^{2x} + 2ce^{3x} = y + (y - ae^x) + ce^{3x}$

$y' = 2y + ce^{3x} - ae^x \color{#20A900}{ . . . . . . . . . (2)}$

$y'' = 2y' + 3ce^{3x} - ae^x = 2y' + (ce^{3x} - ae^x) + 2ce^{3x}$

Using $\color{#20A900}{(2)}$

$y'' = 2y' + (y' - 2y) + 2ce^{3x}$

$y'' = 3y' - 2y + 2ce^{3x} \color{#20A900}{ . . . . . . . . . . (3)}$

Using $\color{#20A900}{(3)}$

$\Rightarrow y''' = 3y'' - 2y' + 6ce^{3x} = 3y'' - 2y' + 3(2ce^{3x})$

$y''' - 3y'' + 2y' = 3(y'' - 3y' + 2y) = 3y'' - 9y' + 6y$

$\boxed{y''' - 6y'' + 11y' - 6y = 0}$

$\large \boxed{\color{#3D99F6}{\dfrac{b\times c}{a \times d}} = \huge \color{#D61F06}{11}}$