# Thread:

1. ## exponential differentiation

Hi, I need to show that y=(Ax+B).e^3x is a solution to y''-6y'+9y=0
I'm getting

y'=e^3x(3Ax+3B+A)

y''=3e^3x(2A+3Ax+3B)

When I sub these values back into the condition above its not working out.

Thank you

2. Originally Posted by slaypullingcat
Hi, I need to show that y=(Ax+B).e^3x is a solution to y''-6y'+9y=0
I'm getting

y'=e^3x(3Ax+3B+A)

y''=3e^3x(2A+3Ax+3B)

When I sub these values back into the condition above its not working out.

Thank you
Show some details - I got it to work!

3. Your calculations for $y'$ and $y''$ are correct. I derived:

$\begin{aligned}<br /> y''-6y'+9y&=(9Ax+9B+6A)e^{3x}\\<br /> &-(18Ax+18B+6A)e^{3x}\\<br /> &+(9Ax+9B)e^{3x}\\<br /> &=0.<br /> \end{aligned}$

4. Your derivatives are correct, and the proposed y is indeed a solution. Check your calculations again (hint: factor out $e^{3x}$).

5. Did it it work with the first and second derivative values I gave? Or did I differentiate incorrectly?

6. Originally Posted by slaypullingcat
Did it it work with the first and second derivative values I gave? Or did I differentiate incorrectly?
Like the others have said - the derivatives are correct. If you show some work, maybe we can show you were you messed up
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