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

Given that
`y_(p(t))=ln((1+2t)), \ \ \ t\ \textgreater \ -(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)= (2(1+6t))/((1+2t)^2)+3ln((1+2t))}`

Answer 2

The given particular solution is

`y_(p)(t) = ln(1 + 2t) - (1)/(2)`

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

Obtain the derivatives of the particular solution.

`y_(p)^(') = (2)/(1+2t) \\\\ y_(p)^('') = (-4)/((1+2t)^(2))`

Because the particular solution should satisfy the ODE, therefore


`g(t)=- (4)/((1+2t)^(2)) +3 \, ln(1+2t)- (3)/(2)`

Answer:
`g(t) = - (4)/((1+2t)^(2)) + 3 \, ln(1+2t) - (3)/(2)`