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?