Doing some exam revision and have been stumped by this; the question asks you to find the recurrence relation satisfied by the 
coefficients.
Attempt at solution:
I have already found that there is a regular singular point at x=0. And by using the indicial equation I know the ODE has a general solution of the form:
I've tried subbing this stuff into the ODE:
but can't figure out where to go from there :( any help would be great
You put all of these three to the starting equation:
$y=\sum\limits_{n=0}^\infty a_nx^{n+\lambda}$
$y'=\sum\limits_{n=0}^\infty a_n (n+\lambda)x^{n+\lambda-1}$
$y''=\sum\limits_{n=0}^\infty a_n (n+\lambda)(n+\lambda-1)x^{n+\lambda-2}$
Now we have:
$2x^2y''=\sum\limits_{n=0}^\infty 2a_n (n+\lambda)(n+\lambda-1)x^{n+\lambda}$
$7xy'=\sum\limits_{n=0}^\infty 7a_n (n+\lambda)x^{n+\lambda}$
$7x^2y'=\sum\limits_{n=0}^\infty 7a_n (n+\lambda)x^{n+\lambda+1}$
$-3y=\sum\limits_{n=0}^\infty -3a_nx^{n+\lambda}$
Now we put all of this in the starting equation and we get:
$\sum\limits_{n=0}^\infty a_n x^{n+\lambda} [2(n+\lambda)(n+\lambda-1) +7(n+\lambda)-3]+ \sum\limits_{n=0}^\infty 7a_n (n+\lambda)x^{n+\lambda+1}=0$
Now to get to the indical equation we need to collect the sums with a use of summation index substitution.
$a_0x^\lambda [2\lambda(\lambda-1) +7\lambda-3] +\sum\limits_{n=1}^\infty a_n x^{n+\lambda} [2(n+\lambda)(n+\lambda-1) +7(n+\lambda)-3]+ \sum\limits_{n=0}^\infty 7a_n (n+\lambda)x^{n+\lambda+1}=0$
$a_0x^\lambda [2\lambda^2+5\lambda-3] +\sum\limits_{n=1}^\infty a_n x^{n+\lambda} [2(n+\lambda)(n+\lambda-1) +7(n+\lambda)-3]+ \sum\limits_{n=0}^\infty 7a_n (n+\lambda)x^{n+\lambda+1}=0$
Now use the substitution $n=m+1$ and get:
$a_0x^\lambda [2\lambda^2+5\lambda-3] +\sum\limits_{m=0}^\infty a_{m+1} x^{m+1+\lambda} [2(m+1+\lambda)(m+\lambda) +7(m+1+\lambda)-3]+ \sum\limits_{n=0}^\infty 7a_n (n+\lambda)x^{n+\lambda+1}=0$
Now the two sums are analog so we can put them in one:
$a_0x^\lambda [2\lambda^2+5\lambda-3] +\sum\limits_{m=0}^\infty [a_{n+1} x^{n+1+\lambda} [2(n+1+\lambda)(n+\lambda) +7(n+1+\lambda)-3]+ 7a_n (n+\lambda)x^{n+\lambda+1}]=0$
$a_0x^\lambda [2\lambda^2+5\lambda-3] +\sum\limits_{m=0}^\infty x^{n+1+\lambda} [a_{n+1} [2(n+1+\lambda)(n+\lambda) +7(n+1+\lambda)-3]+ 7a_n (n+\lambda)]=0$
Now solve for $\lambda$:
$2\lambda^2+5\lambda-3=0$
$\lambda_1=-6;\lambda_2=1$
And you have your indical equation as:
$a_{n+1} [2(n+1+\lambda)(n+\lambda) +7(n+1+\lambda)-3]+ 7a_n (n+\lambda)=0$
Substitute $\lambda$s and that should be it...
For example, for $\lambda_1=-6$ we have:
$a_{n+1} [2(n-5)(n-6) +7(n-5)-3]+ 7a_n (n-6)=0$
$a_{n+1} [2n^2-22 n+60 +7n-35-3]+ 7a_n (n-6)=0$
$a_{n+1} [2n^2-15n+22]+ 7a_n (n-6)=0$
$a_{n+1} =\frac {7 (n-6)}{2n^2-15n+22}a_n$
Now we have:
$a_{n+1} =\frac {14 (n-6)}{(2n+11)(n-2)}a_n$
Analog procedure for $\lambda_2$...
$a_{n+1} [2(n+2)(n+1) +7(n+2)-3]+ 7a_n (n+1)=0$
$a_{n+1} [2n^2+6n+4+7n+14-3]+ 7a_n (n+1)=0$
$a_{n+1} [2n^2+13n+15]+ 7a_n (n+1)=0$
$a_{n+1} =\frac{7a_n (n+1)}{2n^2+13n+15}$