Question

The function y_p(t) = \ln(1 + 2 t), \ t > -\frac{1}{2}, is a particular solution to the differential equation y^{\,\prime\prime} + 3 y = g(t). find g(t)

Answer 1

The given particular solution is
$y_{p}(t) = ln(1 + 2t) - \frac{1}{2}$

The ODE is
y'' + 3y = g(t)

Obtain the derivatives of the particular solution.
$y_{p}^{'} = \frac{2}{1+2t} \\\\ y_{p}^{''} = \frac{-4}{(1+2t)^{2}}$

Because the particular solution should satisfy the ODE, therefore

$g(t)=- \frac{4}{(1+2t)^{2}} +3 \, ln(1+2t)- \frac{3}{2}$

Answer:  $g(t) = - \frac{4}{(1+2t)^{2}} + 3 \, ln(1+2t) - \frac{3}{2}$

Answer 2

Given that $y_{p(t)}=\ln{(1+2t)}, \ \ \ t\ \textgreater \ -\frac{1}{2}$ is a particular solution to the differential equation $y^{\prime\prime}+3y=g(t)$

Then

 $[\ln{(1+2t)}]^{\prime\prime}+3[\ln{(1+2t)}]=g(t) \\ \\ \Rightarrow\bold{g(t)= \frac{2(1+6t)}{(1+2t)^2}+3\ln{(1+2t)}}$