Permalink Submitted by Jhun Vert on September 21, 2019 - 8:54am.

Do you mean eliminate c_{1}, c_{2}, and c_{3} from this equation: y = c_{1}e^{5x} + c_{2}x + c_{3}? If so, this is my solution to it:
$y = c_1 e^{5x} + c_2 x + c_3$

$y' = 5c_1 e^{5x} + c_2$

$y'' = 25c_1 e^{5x}$

$y''' = 125c_1 e^{5x}$

$y''' - 5y'' = 0$ ← this is my answer.

Do you have answer key to this problem? If you do, update us.

The second to the last line is done by taking a derivative. The 4th line was from the division of $e^{5x}$ so that the constant is isolated at the right hand side for it to be eliminated immediately after taking a derivative.

## Do you mean eliminate c1, c2,

Do you mean eliminate

c_{1},c_{2}, andc_{3}from this equation:y=c_{1}e^{5x}+c_{2}x+c_{3}? If so, this is my solution to it:$y = c_1 e^{5x} + c_2 x + c_3$

$y' = 5c_1 e^{5x} + c_2$

$y'' = 25c_1 e^{5x}$

$y''' = 125c_1 e^{5x}$

$y''' - 5y'' = 0$ ← this is my answer.

Do you have answer key to this problem? If you do, update us.

## $$ \begin{eqnarray}

$$ \begin{eqnarray}

y &=& c_1 e^{5x} + c_2 x + c_3\\

y' &=& 5c_1e^{5x} + c_2\\

y'' &=& 25c_1e^{5x}\\

e^{-5x} y'' &=& 25c_1\\

e^{-5x} y''' - 5e^{-5x} y'' &=& 0 \\

y''' - 5y'' &=& 0

\end{eqnarray} $$

The second to the last line is done by taking a derivative. The 4th line was from the division of $e^{5x}$ so that the constant is isolated at the right hand side for it to be eliminated immediately after taking a derivative.

My answer is also $$\boxed{y''' - 5y'' = 0}$$

## Add new comment