# Simple DE problem

• March 29th 2011, 09:10 PM
Vamz
Simple DE problem
$y^{\prime\prime} - 5 y^{\prime} + 0 y=0$

To find the general solution, I just write the auxiliary expression:
$R^2-5R=0$
The solutions for R is 5,0

Therefore, my general solution should be:
$
Y=Ae^{5x}+B
$

But y&#39;&#39;-5y&#39;&#43;0y&#61;0 - Wolfram|Alpha

says otherwise. What am i missing here?
• March 29th 2011, 09:16 PM
pickslides
Quote:

Originally Posted by Vamz
What am i missing here?

Nothing, your solution is fine, just find y' and y'' from your solution and check it in the equation.

Wolfram's solution is also fine.

They are actually the same if you make $\displaystyle A = \frac{1}{5}c_1$