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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

enter image description here

Can anyone see where I've gone wrong?

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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.

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  • $\begingroup$ Thanks, that makes sense! $\endgroup$ – Boris Grunwald Nov 19 '18 at 21:46
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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.

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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

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

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