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
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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!
Your calculations for $\displaystyle y'$ and $\displaystyle y''$ are correct. I derived: $\displaystyle \begin{aligned} y''-6y'+9y&=(9Ax+9B+6A)e^{3x}\\ &-(18Ax+18B+6A)e^{3x}\\ &+(9Ax+9B)e^{3x}\\ &=0. \end{aligned}$
Your derivatives are correct, and the proposed y is indeed a solution. Check your calculations again (hint: factor out $\displaystyle e^{3x}$).
Did it it work with the first and second derivative values I gave? Or did I differentiate incorrectly?
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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