--Short Explanation:
I have to say I am going crazy with this problem as it does not give me the same as the suggested solution in the book:
Problem:
$y''-7y'+10y=9\cos{t}+7\sin{t}$
$y(0)=5$, $y'(0)=-4$
--Long Explanation
So, I've been trying over and over, and then I surrendered and tried finding the solution online, I did find a solution on a website called openstudy.com, this solution gives exactly the same as in the book answers but I am just not following it, here is what I found:
\begin{align} y''-7y'+10y &=9\cos t+7\sin t\qquad y(0)=5, y'(0)=-4\\ s^2Y-sy(0)-y'(0)-7(sY-y(0))+10Y &=\frac{9s}{s^2+1}+\frac7{s^2+1}\\ s^2Y-7sY+10Y-5s+4+35 &=\frac{9s+7}{s^2+1}\\ (s^2-7s+10)Y(s) &=\frac{9s+7}{s^2+1}+5s-39\\ (s^2-7s+10)Y(s) &=\frac{9s+7+5s(s^2+1)-39(s^2+1)}{s^2+1}\\ (s^2-7s+10)Y(s) &=\frac{9s+7+5s^3+1+5s-39s^2-39}{s^2+1}\\ (s-5)(s-2)Y(s) &=\frac{5s^3-39s^2+14s-32}{s^2+1}\\ Y(s) &=\frac{5s^3-39s^2+14s-32}{(s-5)(s-2)(s^2+1)} \end{align}
\begin{align} Y(s) &=\frac{5s^3-39s^2+14s-32}{(s-5)(s-2)(s^2+1)}=\frac{A}{s-5}+\frac{B}{s-2}+\frac{Cs+D}{s^2+1} \\ & A(s-2)(s^2+1)+B(s-5)(s^2+1)+(Cs+D)(s-2)(s-5) \\ &=A(s^3-2s^2+s-2)+B(s^3-5s^2+s-5)+(Cs+D)(s^2-7s+10) \\ &=(A+B)s^3+(-2A-5B)s^2+(A+B)s+(-2A-5B)+(Cs^3+(-7C+D)s^2+(10C-7D)s+10D) \\ &=(A+B+C)s^3-(2A+5B+7C-D)s^2+(A+B+10C-7D)s-(2A+5B+10D) \end{align}
hence we get:$$A+B+C=5\\2A+5B+7C-D=39\\A+B+10C-7D=14\\2A+5B+10D=32$$taking the difference between the second and fourth we get:$$7C-11D=7$$and the first and third gets us:$$9C-7D=9$$eliminating between these two yields (C=1,D=0) and from this we find:$$A+B=4\\2A+5B=32$$hence (A=-4,B=8) and ultimately we conclude:$$Y(s)=-\frac4{s-5}+\frac8{s-2}+\frac{s}{s^2+1}$$
--What I do not understand
Just right before "hence we get:"
\begin{align} & (A+B)s^3+(-2A-5B)s^2+(A+B)s+(-2A-5B)+(Cs^3+(-7C+D)s^2+(10C-7D)s+10D) \\ &=(A+B+C)s^3-(2A+5B+7C-D)s^2 +(A+B+10C-7D)s -(2A+5B+10D) \end{align}
How the heck did that $+10D$ just turned into a $-10D$?, am I just forgetting something?.
Now if this solution is wrong, then how do I go into obtaining the same result, I obtain something completely different. Is the book wrong?
Thank you so much, really.
Ok ... so put the sign back in $2A + 5B - 10D = 32$ and follow the algebra through. The result remains unchanged because $D = 0$, the only value for which $D = -D$.
Hence this 'published' solution is incorrect in that detail, even though the final answer stands.