How do I solve $y''+y'+7y=t$? How do I solve $y''+y'+7y=t$ where $y(0)=0$ and $y'(0)=0$ $(t\geq 0)$?
I tried to solve this by Laplace transformation, but I couldn't find the inverse of $1/(s^2(s^2+s+7))$.
How would I solve this?

You have the Laplace transform, so lets do a partial fraction decomposition. We want to find $a,b,c,d$ such that $$\frac{1}{s^2(s^2 + s +7)} = \frac{a}{s} + \frac{b}{s^2} + \frac{cs + d}{s^2 + s + 7}.$$
So we have
$$as(s^2 + s +7) + b(s^2 + s +7) + cs^3 + ds^2 = 1.$$
Collecting coefficients we have
$$ (a+c)s^3 + (a+b+d)s^2 + (7a+b)s + 7b =1$$
So after working your way through this, you should find, $$a = -\frac{1}{49},b = \frac{1}{7},  d= -\frac{6}{49}, c = \frac{1}{49}.$$
The final trick is to complete the square so that you have
$$ s^2+s +7 = \left(s+\frac{1}{2}\right)^2 +\left(\frac{3\sqrt{3}}{2}\right)^2.$$
So
$$\frac{1}{s^2(s^2 + s +7)} = \frac{1}{49}\left(\frac{-1}{s} + \frac{7}{s^2} + \frac{s+\frac{1}{2} }{\left(s+\frac{1}{2}\right)^2 +\frac{27}{4}}+\frac{13}{3\sqrt{3}}\frac{\frac{3\sqrt{3}}{2}}{\left(s+\frac{1}{2}\right)^2 +\frac{27}{4}}\right).$$
From this point the solution can be read off tables.
Edit:
I'd like to add some insight into wythagoras' answer.
Let's suppose you have some function $y$. Suppose then you differentiate it a couple of times and you end up with a polynomial. Then it follows that $y$ had to be a polynomial to start with. Remember that differentiating polynomials gives you another polynomial of one less order. Hence, if i add a polynomial to it's derivatives, then what comes out the other end must be the same order as what i started with. So when i have $y'' + y'+7y =t$ then i know that i must have a polynomial, and that the highest order of that polynomial must be of order $t$. So we are well justified assuming that the particular solution has the form $y_p= bt +a$.
