Originally Posted by

**Lancet** I solved an ODE, but I'm finding a discrepancy between the answer I came up with and what Wolfram Alpha says is the answer.

Can anyone tell me if I made a mistake, or if my answer is also valid?

$\displaystyle y'' - 5y' + 6y = 2e^{3t} + cos(t)$

$\displaystyle r^2 - 5r + 6 = 0$

$\displaystyle r = 2, 3$

$\displaystyle y_h = C_1e^{2t} + C_2e^{3t}$

$\displaystyle y_1 = e^{2t}$

$\displaystyle y_2 = e^{3t}$

$\displaystyle W(y_1, y_2) = e^{5t}$

$\displaystyle g(t) = 2e^{3t} + cos(t)$

$\displaystyle u = \int -\frac{(2e^{3t} + cos(t))e^{3t}}{e^{5t}} dt$

$\displaystyle u = -2e^t - \frac{1}{5}e^{-2t}(sin(t) - 2cos(t)) + C_1$

$\displaystyle v = \frac{(2e^{3t} + cos(t))2e^{2t}}{e^{5t}}$

$\displaystyle v = 4t + \frac{1}{5}e^{-3t}(sin(t) - 3cos(t)) + C_2$

$\displaystyle y = uy_1 + vy_2$

$\displaystyle y = e^{2t}[-2e^t - \frac{1}{5}e^{-2t}(sin(t) - 2cos(t)) + C_1] + e^{3t}[4t + \frac{1}{5}e^{-3t}(sin(t) - 3cos(t)) + C_2]$

$\displaystyle y = C_1e^{2t} + C_2e^{3t} - 2e^{3t} + 4te^{3t} - \frac{1}{5}cos(t)$

Wolfram Alpha says the answer is this:

$\displaystyle y = C_1e^{2t} + C_2e^{3t} + 2te^{3t} - \frac{1}{10}sin(t) + \frac{1}{10}cos(t)$

Did I make a mistake?