Question A Let the answer to question a be of the form

\(\displaystyle{a}{y}{''}+{b}{y}'+{c}{y}={0}\)

Then we want \(\displaystyle{a}{\left(-{3}\right)}^{{2}}-{3}{b}+{c}={9}{a}-{3}{b}+{c}={0}\) (1)

And more over, since it is a repeated root we need

\(\displaystyle-{\frac{{{b}}}{{{2}{a}}}}=-{3}\)

Substituting \(\displaystyle{b}={6}{a}\) into (1) we get

\(\displaystyle{9}{a}-{18}{a}+{c}={0}\Rightarrow={9}{a}\)

Thus if we let a be our free variable, all solutions to question a are of the form

\(\displaystyle{a}{y}{''}+{6}{a}{y}'+{9}{a}{y}={0}\)

Question B For question b, you need the characteristic polynomial to have the roots \(\displaystyle{3}\pm{i}\). Thus the characteristic polynomial must be of the form

\(\displaystyle{a}{\left({x}-{\left[{3}+{i}\right]}\right)}{\left({x}-{\left[{3}-{i}\right]}\right)}={a}{x}^{{2}}-{6}{a}{x}+{10}{a}\)

Thus solutions to question b are of the form

\(\displaystyle{a}{y}{''}-{6}{a}{y}'+{10}{a}{y}={0}\)

Note For your solution to question a you got \(\displaystyle{y}{''}-{6}{y}'+{9}{y}={0}\) but going back and checking you should get \(\displaystyle+{6}{y}'\) not \(\displaystyle-{6}{y}'\). Otherwise your solutions are of these forms with \(\displaystyle{a}={1}\) but a can be any real number. It is our free variable.

\(\displaystyle{a}{y}{''}+{b}{y}'+{c}{y}={0}\)

Then we want \(\displaystyle{a}{\left(-{3}\right)}^{{2}}-{3}{b}+{c}={9}{a}-{3}{b}+{c}={0}\) (1)

And more over, since it is a repeated root we need

\(\displaystyle-{\frac{{{b}}}{{{2}{a}}}}=-{3}\)

Substituting \(\displaystyle{b}={6}{a}\) into (1) we get

\(\displaystyle{9}{a}-{18}{a}+{c}={0}\Rightarrow={9}{a}\)

Thus if we let a be our free variable, all solutions to question a are of the form

\(\displaystyle{a}{y}{''}+{6}{a}{y}'+{9}{a}{y}={0}\)

Question B For question b, you need the characteristic polynomial to have the roots \(\displaystyle{3}\pm{i}\). Thus the characteristic polynomial must be of the form

\(\displaystyle{a}{\left({x}-{\left[{3}+{i}\right]}\right)}{\left({x}-{\left[{3}-{i}\right]}\right)}={a}{x}^{{2}}-{6}{a}{x}+{10}{a}\)

Thus solutions to question b are of the form

\(\displaystyle{a}{y}{''}-{6}{a}{y}'+{10}{a}{y}={0}\)

Note For your solution to question a you got \(\displaystyle{y}{''}-{6}{y}'+{9}{y}={0}\) but going back and checking you should get \(\displaystyle+{6}{y}'\) not \(\displaystyle-{6}{y}'\). Otherwise your solutions are of these forms with \(\displaystyle{a}={1}\) but a can be any real number. It is our free variable.