First, you see, I am sure, that this is a second order differential equation. Such equations will typically have two independent solutions and, since this is a linear differential equation, the general solution is linear combination of those two solutions, requiring two conditions the solution must satisfy in addition to the differential equation. The condition that "\(\displaystyle c_0= 1\)" is one condition but because there is only one condition, we will have to leave \(\displaystyle c_1\) as an undetermined constant.
The \(\displaystyle c_n\) in the problem is the \(\displaystyle a_n\) in your computation. So your formula is \(\displaystyle a_n= \frac{a_{n-1}}{(r+ n)(r+n-1)+ 5(r+ n)+ 4}\). You are given that \(\displaystyle c_0= a_0= 1\) so \(\displaystyle a_1= \frac{a_0}{r(r-1)+ 5r+ 4}\). IF you could use that to calculate \(\displaystyle a_1\), all other coefficients would follow from that and you would have a single solution. In order that we NOT be able to do that, we must have the denominator, \(\displaystyle r(r- 1)+ 5r+ 4= r^2+ 4r+ 4= (r+2)^2= 0\). That is, r= -2. With r= 0, we can choose \(\displaystyle a_1= c_1\) ourselves. Then other values for \(\displaystyle a_n= c_n\) you can get from \(\displaystyle a_n= \frac{a_{n-1}}{(5+ n)(5+n-1)+ 5(5+ n)+ 4}= \frac{a_{n-1}}{(5+n)(4+n)+ 25+ 5n+ 4}= \frac{a_{n-1}}{n^2+ 14n+ 49}\).
