Cfrgtkky

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?

You have not gone wrong anywhere! It is just a question concerning the choice of the arbitrary constants $c_1,c_2$ and $C_1,C_2$ respectively. Therefore lets just take a brief look at the two given solutions

$$begin{align}
y_1(t)&=c_1+c_2e^{-7t}-2t\
y_2(t)&=C_2+frac{-C_1}{7}e^{-7t}-2t
end{align}$$

We can agree one the fact that the particular solution $-2t$ is the same in both $y_1(t)$ and $y_2(t)$. Therefore we do not have to worry about this term further. Now we can argue that $c_1=C_2$ since these are both arbitrary constants which have to be defined later one. Following a similiar agrumentation we can moreover say that $c_2=frac{-C_1}{7}$ hence there is no restriction regarding their value.

Therefore Maple aswell as you by yourself provided the exact same solution beside the choice of some arbitrary constants.

You went wrong when you thought what Maple wrote is different from your solution in any significant way.

Maple has swapped the roles of $c_1$ and $c_2$ compared to you. And Maple's $c_1$ is seven times larger than your $c_2$ and has opposite sign, but since the constants are arbitrary anyways this doesn't matter. So you and Maple describe the exact same collection of functions.

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

$$y=c_0e^{-7t}-2t+c_1.$$