Let's classify this differential equation. It is

1.linear. I.e. there are no y's with squares or inside cosines or something.

2.second-order. I.e. the highest derivative of y is 2.

3.homogeneous. I.e. there are no terms without a y in it.

The first and third of these properties tell us that this is an Euler-Cauchy equation. That's cool because it means we know there's a trick to solving this. What we need to do is try out a solution of the form \(\displaystyle{y}={x}^{{k}}\) and see what happens. It won't end up being the correct solution, but it should (hopefully) help us figure out the correct solution.

So let's do that. Firstly, we need to calculate y'' when \(\displaystyle{y}={x}^{{k}}\):

\(\displaystyle{y}'={k}{x}^{{{k}-{1}}}\)

\(\displaystyle{y}{''}={k}{\left({k}-{1}\right)}{x}^{{{k}-{2}}}\)

Now let's substitute these values into the differential equation.

\(\displaystyle{4}{x}^{{2}}{\left[{k}{\left({k}-{1}\right)}{x}^{{{k}-{2}}}\right]}+{17}{\left[{x}^{{k}}\right]}={0}\)

\(\displaystyle{\left({4}{k}^{{2}}-{4}{k}+{17}\right)}{x}^{{k}}={0}\)

\(\displaystyle{4}{k}^{{2}}-{4}{k}+{17}={0}\)

(everywhere where x does not equal 0)

So now we just solve this using whatever your favorite method for solving quadratic equations is to get the solutions

\(\displaystyle{k}=\frac{{1}}{{2}}\pm{2}{i}\)

Now from the section in your book on Euler-Cauchy equations, we should know that the general solution to this DE is

\(\displaystyle{y}{\left({x}\right)}={A}{x}^{{\frac{{1}}{{2}}}}{\cos{{\left({2}{\ln{{\left({x}\right)}}}\right)}}}+{B}{x}^{{\frac{{1}}{{2}}}}{\sin{{\left({2}{\ln{{\left({x}\right)}}}\right)}}}\)

However, that's not the end of the problem because we still need to use those initial values. Let's plug them in and see if we can figure out what A and B have to be.

\(\displaystyle{y}{\left({1}\right)}={A}{\left({1}\right)}^{{\frac{{1}}{{2}}}}{\cos{{\left({2}\cdot{0}\right)}}}+{B}{\left({1}\right)}^{{\frac{{1}}{{2}}}}{\sin{{\left({2}\cdot{0}\right)}}}=-{1}\)

\(\displaystyle\Rightarrow{A}=-{1}\)

\(\displaystyle{y}'{\left({x}\right)}=\frac{{{A}+{4}{B}}}{{{2}\sqrt{{x}}}}{\cos{{\left({2}{\ln{{\left({x}\right)}}}\right)}}}+\frac{{-{4}{A}+{B}}}{{{2}\sqrt{{x}}}}{\sin{{\left({2}{\ln{{\left({x}\right)}}}\right)}}}\)

\(\displaystyle{y}'{\left({1}\right)}=\frac{{{\left(-{1}\right)}+{4}{B}}}{{2}}{\left({1}\right)}+{0}=-\frac{{1}}{{2}}\)

\(\displaystyle\Rightarrow{B}={0}\)

Hence your final solution is

\(\displaystyle{y}{\left({x}\right)}=-{x}^{{\frac{{1}}{{2}}}}{\cos{{\left({2}{\ln{{\left({x}\right)}}}\right)}}}\)

1.linear. I.e. there are no y's with squares or inside cosines or something.

2.second-order. I.e. the highest derivative of y is 2.

3.homogeneous. I.e. there are no terms without a y in it.

The first and third of these properties tell us that this is an Euler-Cauchy equation. That's cool because it means we know there's a trick to solving this. What we need to do is try out a solution of the form \(\displaystyle{y}={x}^{{k}}\) and see what happens. It won't end up being the correct solution, but it should (hopefully) help us figure out the correct solution.

So let's do that. Firstly, we need to calculate y'' when \(\displaystyle{y}={x}^{{k}}\):

\(\displaystyle{y}'={k}{x}^{{{k}-{1}}}\)

\(\displaystyle{y}{''}={k}{\left({k}-{1}\right)}{x}^{{{k}-{2}}}\)

Now let's substitute these values into the differential equation.

\(\displaystyle{4}{x}^{{2}}{\left[{k}{\left({k}-{1}\right)}{x}^{{{k}-{2}}}\right]}+{17}{\left[{x}^{{k}}\right]}={0}\)

\(\displaystyle{\left({4}{k}^{{2}}-{4}{k}+{17}\right)}{x}^{{k}}={0}\)

\(\displaystyle{4}{k}^{{2}}-{4}{k}+{17}={0}\)

(everywhere where x does not equal 0)

So now we just solve this using whatever your favorite method for solving quadratic equations is to get the solutions

\(\displaystyle{k}=\frac{{1}}{{2}}\pm{2}{i}\)

Now from the section in your book on Euler-Cauchy equations, we should know that the general solution to this DE is

\(\displaystyle{y}{\left({x}\right)}={A}{x}^{{\frac{{1}}{{2}}}}{\cos{{\left({2}{\ln{{\left({x}\right)}}}\right)}}}+{B}{x}^{{\frac{{1}}{{2}}}}{\sin{{\left({2}{\ln{{\left({x}\right)}}}\right)}}}\)

However, that's not the end of the problem because we still need to use those initial values. Let's plug them in and see if we can figure out what A and B have to be.

\(\displaystyle{y}{\left({1}\right)}={A}{\left({1}\right)}^{{\frac{{1}}{{2}}}}{\cos{{\left({2}\cdot{0}\right)}}}+{B}{\left({1}\right)}^{{\frac{{1}}{{2}}}}{\sin{{\left({2}\cdot{0}\right)}}}=-{1}\)

\(\displaystyle\Rightarrow{A}=-{1}\)

\(\displaystyle{y}'{\left({x}\right)}=\frac{{{A}+{4}{B}}}{{{2}\sqrt{{x}}}}{\cos{{\left({2}{\ln{{\left({x}\right)}}}\right)}}}+\frac{{-{4}{A}+{B}}}{{{2}\sqrt{{x}}}}{\sin{{\left({2}{\ln{{\left({x}\right)}}}\right)}}}\)

\(\displaystyle{y}'{\left({1}\right)}=\frac{{{\left(-{1}\right)}+{4}{B}}}{{2}}{\left({1}\right)}+{0}=-\frac{{1}}{{2}}\)

\(\displaystyle\Rightarrow{B}={0}\)

Hence your final solution is

\(\displaystyle{y}{\left({x}\right)}=-{x}^{{\frac{{1}}{{2}}}}{\cos{{\left({2}{\ln{{\left({x}\right)}}}\right)}}}\)