Answer:
The solution of the differential equation is
.
Step-by-step explanation:
The first step is to take Laplace transform in both sides of the differential equation. As usual, we denote the Laplace transform of
as
. Then,
 = L[36t](s)](https://tex.z-dn.net/?f=L%5By%27%2B3y%5D%28s%29%20%3D%20L%5B36t%5D%28s%29)
+3L[y](s) = 36L[t](s)](https://tex.z-dn.net/?f=L%5By%27%5D%28s%29%2B3L%5By%5D%28s%29%20%3D%2036L%5Bt%5D%28s%29)

In the last step we use that
and
.
Notice that our differential equations becomes an algebraic equation for
, which is more simple to solve.
In the expression we have obtained, we can write
in terms of
:
which is equivalent to
.
Now, we make a partial fraction decomposition for the term
. Thus,
.
Substituting the above value into the expression for
we get
) in both hands of the above expression. Recall that
. So,
](https://tex.z-dn.net/?f=y%28t%29%20%3D%20L%5E%7B-1%7D%5Cleft%5B-%5Cfrac%7B4%7D%7Bs%7D%20%2B%20%5Cfrac%7B12%7D%7Bs%5E2%7D%20%2B%20%5Cfrac%7B10%7D%7Bs%2B3%7D%5Cright%5D%28t%29%20)
 + 12L^{-1}\left[\frac{1}{s^2}\right](t) + 10L^{-1}\left[\frac{1}{s+3}\right](t)](https://tex.z-dn.net/?f=y%28t%29%20%3D%20-4L%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs%7D%5Cright%5D%28t%29%20%2B%2012L%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs%5E2%7D%5Cright%5D%28t%29%20%2B%2010L%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs%2B3%7D%5Cright%5D%28t%29%20)
.
To obtain this we have used the following identities that can be found in any table of Laplace transforms
 = 1](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs%7D%5Cright%5D%28t%29%20%3D%201)
 = t](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs%5E2%7D%5Cright%5D%28t%29%20%3D%20t)
 = e^{-3t}](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs%2B3%7D%5Cright%5D%28t%29%20%3D%20e%5E%7B-3t%7D)