Differential equation: Eliminate the arbitrary constant from $y=c_1e^{5x}+c_2x+c_3$

Submitted by Awing on

y=c1e^5x+c2x+c3

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Elimination of Arbitrary Constant

Jhun Vert Sat, 09/21/2019 - 08:54

Do you mean eliminate c1, c2, and c3 from this equation: y = c1e5x + c2x + c3? 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.

Infinitesimal Mon, 03/02/2020 - 18:42

$$ \begin{eqnarray}
y &=& c_1 e^{5x} + c_2 x + c_3\\
y' &=& 5c_1e^{5x} + c_2\\
y'' &=& 25c_1e^{5x}\\
e^{-5x} y'' &=& 25c_1\\
e^{-5x} y''' - 5e^{-5x} y'' &=& 0 \\
y''' - 5y'' &=& 0
\end{eqnarray} $$

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 $$\boxed{y''' - 5y'' = 0}$$