<s>[INST] Using inverse Laplace transform to solve differential equation The differential equation is as follows- 
$$\frac{d^2 x}{dt^2} + 5 \frac{dx}{dt} + 6x = e^t $$
I use laplace transform to make it to become - $$X(s) = \frac{1}{(s-1)(s+3)(s+2)}$$
where $X(s)$ is the Laplace transform of $X(t)$
So now I am trying to find $X(t)$ using inverse transform.
From partial fractions- 
$X(s) = \frac{1}{(s-1)(s+3)(s+2)} = \frac{A}{s-1} + \frac{B}{s+3} + \frac{C}{s+2} $
Numerator - $ 1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3) $ 
I am stuck from here on how to carry on this partial fraction 
Can I sub all s values to be 0 ? 
For example 
$1 = A(0+3)(0+2)$
$1= B(0-1)(0+2) $
$1 = C (0-1)(0+3) $ 
 [/INST] $ s=0$ is not the best choice
There are three values to assign to $s$ which makes our life very easy.
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
$$ s=1 \implies 12A=1 \implies A=\frac {1}{12}$$
$$ s=-3 \implies 4B=1 \implies B=\frac {1}{4}$$ 
$$ s=-2 \implies -3C=1 \implies C=\frac {-1}{3}$$
Now you proceed with the inverse Laplace Transform.
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