TITLE: Are these wrong signs because of a book error or wrong procedure?
QUESTION [1 upvotes]: I was solving these DEs
$\frac{dx}{dt} = -y+t$
$\frac{dy}{dt} = x-t$
Doing it by elimination they become
$x''+x = t+1$
$y''+y = t-1$
...for $x$ and $y$ respectively. If I solve those I get
$x = c_1\cos(t) + c_2\sin(t) + t + 1$
$y = c_1\cos(t) + c_2\sin(t) + t - 1$
Now, it makes sense to me that the general solution (Cos and Sin part) is the same for both cases, but the solution says
$x = c_1\cos(t) + c_2\sin(t) + t + 1$
$y = c_1\sin(t) - c_2\cos(t) + t - 1$
And this happens with the solutions of a few other problems too, so I'm not quite convinced it's just a print error. Am I doing something wrong?

REPLY [3 votes]: What you are doing wrong is taking the constants $c_1$ and $c_2$ you used to write the "general" solution for $x(t)$ and re-using them to write the general solution for $y(t)$. 
Even though you separated $x$ and $y$ they still have to satisfy your original "mixed" equations. By re-using $c_1$ and $c_2$ you are assuming that if, say through the addition of some initial conditions, we find that the co-efficient on the $sin$ term in $x$ is 3, then the the co-efficient on the $sin$ term of $y$ must also necessarily be 3. Nothing justifies this.
Here's what you should do: 

Separate them, just like you did
Write the general solution for $x(t)$, using $c_1$ and $c_2$
Then write the general solution for $y(t)$, using new co-efficients $c_3$ and $c_4$
Substitute these solutions back into the original equations to see what restrictions this places on the $c$'s. If this yields something complicated all you can do is state this as a requirement for your $x$ and $y$ to be solutions. Or it might be simple and let you re-write some of the $c$'s in terms of others, and you can eliminate those $c$'s completely.

I haven't worked out the details, but I'll bet that in your case the last step will give you $c_1 = c_3$ and $c_2 = -c_4$, so eliminating $c_3$ and $c_4$ will give you the given solution.
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