The given particular solution is
[tex]y_{p}(t) = ln(1 + 2t) - \frac{1}{2} [/tex]
The ODE is
y'' + 3y = g(t)
Obtain the derivatives of the particular solution.
[tex]y_{p}^{'} = \frac{2}{1+2t} \\\\ y_{p}^{''} = \frac{-4}{(1+2t)^{2}} [/tex]
Because the particular solution should satisfy the ODE, therefore
[tex]g(t)=- \frac{4}{(1+2t)^{2}} +3 \, ln(1+2t)- \frac{3}{2} [/tex]
Answer: [tex]g(t) = - \frac{4}{(1+2t)^{2}} + 3 \, ln(1+2t) - \frac{3}{2} [/tex]
