Multiply through by $m$ to give $my'' + dy' + ky = Gm.$ First, find the complementary function by solving the homogeneous equation $my'' + dy' + ky = 0.$ Use the trial function $y(t) = e^{\lambda t}$. We get $(m\lambda^2 + d\lambda +k)e^{\lambda t} = 0.$ Since $e^{\lambda t} > 0$ we need to solve $m\lambda^2 + d\lambda +k = 0$ for $\lambda.$ Whence:
$\lambda_{\pm} = \frac{1}{2m}(-d \pm \sqrt{d^2 - 4mk}) \, .$
The complimentary function is then $y(t) = ae^{\lambda_-} + be^{\lambda_+}$ for any $a$ and $b$. Next, we need to find the particular integral. Since the right hand side of $my'' + dy' + ky = Gm$ is constant, we try $y(t) = \alpha$ where $\alpha$ is a constant. Putting that into $my'' + dy' + ky = Gm$ gives $k\alpha = Gm$ and so $\alpha = Gm/k.$ The general solution is then:
$y(t) = ae^{\lambda_-} + be^{\lambda_+} + \frac{Gm}{k} \, .$
Finally, we impose the initial conditions that $y(0) = t_0$ and $y'(0) = 0.$ You will get two simultaneous equations in $a$ and $b$. Solve them gives:
$a = \frac{m\lambda_+(Gm-kt_0)}{k\sqrt{d^2-4mk}} \, , $
$b = \frac{m\lambda_-(Gm-kt_0)}{k(d^2-4mk)} \, . $
Provided I've not made a silly mistake, I get:
$y(t) = \frac{m}{k}\left[ G + \frac{(Gm-kt_0)}{\sqrt{d^2-4mk}}\left(\lambda_+e^{\lambda_-t} + \lambda_-e^{\lambda_+t}\right) \right] \, . $