Originally Posted by

**petition Edgecombe** I did have $\displaystyle \frac{3}{5} e^t$, but I forgot how I got it, so I changed it (that's why I edited my post).

I see what I did there. I included the 3 in the initial guess so that it looked like

$\displaystyle y_{p2}=3Ae^t$ instead of $\displaystyle y_{p2}=Ae^t$

I never really learned about equating coefficients. I briefly used it in calculus for partial fraction decomposition, that's it.

$\displaystyle y_{p}=At^2+Bt+C$

$\displaystyle y_{p}'=2At+B$

$\displaystyle y_{p}''=2A$

$\displaystyle 2A+4At^2+4Bt+4C=t^2+0t+0c$ Mr F says: This can be re-arranged as 4At^2 + 4Bt + (2A + 4C) = t^2 + 0t + 0. The constant term on the left hand side is what you've failed to see properly ..... See below.

$\displaystyle 4At^2=t^2$

$\displaystyle A=1/4$

$\displaystyle 4Bt=0t$

$\displaystyle B=0$

$\displaystyle C=0c$ Mr F says: Wrong.

This is where I knew the mistake would be. I was hoping you would have realised it while typing your work out. The correct equation is 2A + 4C = 0. Substitute A = 1/4 and solve for C.

$\displaystyle C = 0$ Mr F says: No. C = ....