
Simple DE problem
$\displaystyle y^{\prime\prime}  5 y^{\prime} + 0 y=0$
To find the general solution, I just write the auxiliary expression:
$\displaystyle R^25R=0$
The solutions for R is 5,0
Therefore, my general solution should be:
$\displaystyle
Y=Ae^{5x}+B
$
But y''5y'+0y=0  WolframAlpha
says otherwise. What am i missing here?

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 $\displaystyle \displaystyle A = \frac{1}{5}c_1$