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