I was eliminating the arbitrary constants of the equation $y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x}$.
My work:
I see three arbitrary constants, so I need to differentiate the equation $y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x}$ thrice.
Now differentiating, I get….
$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \space \space \space —–> (1)$$
$$y' = c_1 e^x + 2c_2 e^{2x} + 3c_3 3e^{3x} \space \space \space —–> (2)$$
$$y'' = c_1 e^x + 4c_2 e^{2x} + 9c_3 3e^{3x} \space \space \space —–> (3)$$
$$y''' = c_1 e^x + 8c_2 e^{2x} + 27c_3 3e^{3x} \space \space \space —–> (4) $$
I'm eliminating the $c_1$. Looking at equation's $(1)$ and $(2)$:
$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \space \space \space —–> (1)$$
$$y' = c_1 e^x + 2c_2 e^{2x} + 3c_3 3e^{3x} \space \space \space —–> (2)$$
Then I multiply equation $(2)$ by $-1$ and adding it to equation $(1)$, we get:
$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \space \space \space —–> (1)$$
$$+ (-1)\left(y' = c_1 e^x + 2c_2 e^{2x} + 3c_3 3e^{3x} \right) \space \space \space —–> (2)$$
$————————————–$
$$y -y' = -c_2e^{2x} – 2c_3e^{3x} ———–> (A)$$
I'm eliminating another $c_1$. Looking at equation's $(1)$ and $(3)$:
$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \space \space \space —–> (1)$$
$$y'' = c_1 e^x + 4c_2 e^{2x} + 9c_3 3e^{3x} \space \space \space —–> (3)$$
Then I multiply equation $(3)$ by $-1$ and adding it to equation $(1)$, we get:
$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \space \space \space —–> (1)$$
$$+(-1)\left(y'' = c_1 e^x + 4c_2 e^{2x} + 9c_3 3e^{3x} \right) \space \space \space —–> (3)$$
$————————————–$
$$y -y'' = -3c_2e^{2x} – 8c_3e^{3x} ————> (B)$$
I'm eliminating the last $c_1$. Looking at equation's $(1)$ and $(4)$:
$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \space \space \space —–> (1)$$
$$y''' = c_1 e^x + 8c_2 e^{2x} + 27c_3 3e^{3x} \space \space \space —–> (4) $$
Then I multiply equation $(4)$ by $-1$ and adding it to equation $(1)$, we get:
$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \space \space \space —–> (1)$$
$$+(-1)\left(y''' = c_1 e^x + 8c_2 e^{2x} + 27c_3 3e^{3x} \right) \space \space \space —–> (4)$$
$————————————–$
$$y -y''' = -7c_2e^{2x} – 26c_3e^{3x} ———-> (C)$$
After all those process, I got to eliminate $c_2$. Looking at equations $(A)$ and $(B)$:
$$y -y' = -c_2e^{2x} – 2c_3e^{3x} ———-> (A)$$
$$y -y'' = -3c_2e^{2x} – 8c_3e^{3x} ———–> (B)$$
Then I multiply equation $(A)$ by $-3$ and adding it to equation $(B)$, we get:
$$(-3)\left(y -y' = -c_2e^{2x} – 2c_3e^{3x} \right)————> (-3A)$$
$$+(y -y'' = -3c_2e^{2x} – 8c_3e^{3x}) ———–> (B)$$
$————————————–$
$$-2y+3y'-y'' = -2c_3e^{3x} ————> (\alpha) $$
I got to eliminate another $c_2$ finally. Looking at equations $(A)$ and $(C)$:
$$y -y' = -c_2e^{2x} – 2c_3e^{3x} ———–> (A)$$
$$y -y''' = -7c_2e^{2x} – 26c_3e^{3x} ————> (C)$$
Then I multiply equation $(A)$ by $-7$ and adding it to equation $(C)$, we get:
$$(-7)\left(y -y' = -c_2e^{2x} – 2c_3e^{3x} \right)———–> (-7A)$$
$$+(y -y''' = -7c_2e^{2x} – 26c_3e^{3x}) ———–> (C)$$
$————————————–$
$$-6y+7y'-y''' = -12c_3e^{3x} ————> (\beta) $$
Now getting their $c_3$'s in $(\alpha)$ and $(\beta)$, we get:
for $(\alpha)$ : $$c_3 = \frac{-2y+3y'-y''}{-2e^{3x}}$$
for $(\beta)$ : $$c_3 = \frac{-6y+7y'-y'''}{-12e^{3x}}$$
Then equate their $c_3$'s :
$$\frac{-2y+3y'-y''}{-2e^{3x}} = \frac{-6y+7y'-y'''}{-12e^{3x}}$$
Then….
$$\frac{-2y+3y'-y''}{-2e^{3x}} = \frac{-6y+7y'-y'''}{-12e^{3x}}$$
$$\left( \frac{-2y+3y'-y''}{-2e^{3x}} = \frac{-6y+7y'-y'''}{-12e^{3x}} \right) (12e^{3x})$$
$$(-6)(-2y+3y'-y''= -6y+7y'-y''') $$
$$12y – 18y' + 6y'' = 36y -42y' + 42y''' $$
$$42y''' -6y'' – 24y' + 24y = 0$$
Ultimately, when I eliminate arbitrary constants of the equation $y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x}$, I got the new equation
$$7y''' -y'' -4y' + 4y = 0$$
I thought I got the answer correctly because when I solved it, it was smooth…..but I couldn't verify the true answer.
My question is: Is my answer ($7y''' -y'' -4y' + 4y = 0$) correct?
Best Answer
You can work this out by Gaussian elimination
$$\begin{matrix}y&1&1&1\\y'&1&2&3\\y''&1&4&9\\y'''&1&8&27\\\end{matrix}$$
$$\begin{matrix}y&1&1&1\\y'-y&0&1&2\\y''-y&0&3&8\\y'''-y&0&7&26\\\end{matrix}$$
$$\begin{matrix}y&1&1&1\\y'-y&0&1&2\\y''-3y'+2y'&0&0&2\\y'''-3y''+2y'&0&0&12\\\end{matrix}$$
$$\begin{matrix}y&1&1&1\\y'-y&0&1&2\\y''-3y'+2y&0&0&2\\y'''-6y''+11y'-6y&0&0&0\\\end{matrix}$$
But it is much simpler to consider that the operator $D-a$ eliminates the exponential $e^{ax}$, so that
$$(D-1)(D-2)(D-3)y=(D^3-6D+11D-6)y=0.$$
Indeed, $$(D-1)y=c_1e^x-c_1e^x+2c_2e^{2x}-c_2e^{2x}+3c_3e^{3x}-c_3e^{3x}=c'_2e^{2x}+c'_3e^{3x}$$ $$(D-2)(D-1)y=2c'_2e^{2x}-2c'_2e^{2x}+3c_3e^{3x}-2c_3e^{3x}=c''_3e^{3x}$$
$$(D-3)(D-2)(D-1)y=3c''_2e^{3x}-3c''_2e^{3x}=0.$$