<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] $$X(s) = \frac{1}{(s-1)(s+3)(s+2)}$$
Substitute $k=s+2$
$$F(k) = \frac{1}{(k-3)(k+1)k}$$
$$F(k) = \frac{1}{(k-3)}\left(\frac 1 {(k+1)k} \right)$$
$$F(k) = \frac{1}{(k-3)}\left(\frac 1 k-\frac 1 {k+1} \right)$$
$$F(k) = \frac{1}{(k-3)}\frac 1 k-\frac{1}{(k-3)}\frac 1 {(k+1)} $$
$$F(k) = \frac 13 \left (\frac{1}{(k-3)}-\frac 1 k\right )-\frac 14\left(\frac{1}{(k-3)}-\frac 1 {(k+1)} \right )$$
$$F(k) = \frac 1 {12} \frac{1}{(k-3)}-\frac 1{3k}+\frac 14\frac 1 {(k+1)} $$
Substitute back $k=s+2$
$$X(s)=-\frac 1{3(s+2)}+\frac 1{4(s+3)}+\frac 1 {12(s-1)}$$
$$x(t)=-\frac 1{3}e^{-2t}+\frac 1{4}e^{-3t}+\frac 1 {12}e^{t}$$
It's a bit longer but i find this method easier
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