I need help finding the general solution to the differential equation $y''(t)+7y'(t)=-14$ What I've tried:
I have the inhomogeneous differential equation:
$$y''(t)+7y'(t)=-14$$
I find the particular solution to be on the form $$kt$$
by inserting the particular solution in the equation
$$(kt)''+7(kt)'=-14$$
and isolating for k, I get that:
$$k=-2$$
and therefore the particular solution is
$$y(t)-2t$$
I also need the general solution for the homogenous equation
$$y''(t)+7y'(t)=0$$
by finding the roots of the characteristic polynomial
$$z^2+7z=z(z+7)=0$$
$$z_1=0$$
$$z_2=-7$$
I get the general solution:
$$c_1e^{0t}+c_2e^{-7t}=c_1+c_2e^{-7t}$$
Now, according to my textbook, the general solution of an inhomogeneous differential equation is given by
$$y(t)=y_p(t)+y_{hom}(t)$$
Where $y_p(t)$ is the particular solution and $y_{hom}(t)$ is the general solution to the homogenous equation. Therefore I get the general solution to be
$$y(t)=c_1+c_2e^{-7t}-2t$$
This is not consistent with Maple's result however

Can anyone see where I've gone wrong?

As others said, your solution is right (only the expressions of the constant differ). Here is how I would solve.
First notice that $y$ is missing, and solve for $z:=y'$:
$$z'+7z=-14.$$
An obvious particular solution is the constant $z=-2$. Then with the characteristic polynomial $p+7$, you can immediately write the general solution,
$$z=ce^{-7t}-2.$$
Now to get $y$, you integrate once, and obtain
$$y=c_0e^{-7t}-2t+c_1.$$
