Consider the second linearly independent solution as \(g(t)=v(t),\ f(t)=3ve^{2t}

Obtain derivatives:

\(g'(t)=6ve^{2t}+3e^{2t}v'\)

and

\(g''(t)=12ve^{2t}+6e^{2t}v'+6e^{2t}v'+3e^{2t}v''\)

\(=12ve^{2t}+12e^{2t}v'+3e^{2t}v''\)

Substitute g and its derivative:

\(t(12ve^{2t}+12e^{2t}v'+3e^{2t}v'')-(2t+1)(6ve^{2t}+3e^{2t}v')+2(3ve^{2t})=0\)

\(12tve^{2t}+12te^{2t}v'+3te^{2t}v''-12tve^{2t}-6te^{2t}v'-6ve^{2t}-3e^{2t}v'+6ve^{2t}=0\)

\(3te^{2t}v''+(6t-3)e^{2t}v'=0\)

\(3tv''+(6t-3)v'=0\)

Further follows,

Let \(w=v'\), then

\(3tw'+(6t-3)w=0\)

\(3tw'=-(6t-3)w\)

\(\frac{dw}{w}=\frac{3-6t}{3t}dt\)

\(\frac{dw}{w}=(\frac{1}{t}-2)dt\)

Integrate both sides:

\(\int\frac{dw}{w}=\int(\frac{1}{2}-2)dt\)

\(\ln w=\ln t-2t+C\)

\(w=e^{\ln t-2t+C}=Ate^{-2t}\)

Then, we have

As \(w=v'\), obtain the function v(t):

\(v'=Ate^{-2t}\)

\(v(t)=-\frac{A}{2}te^{-2t}+\int\frac{A}{2}e^{-2t}dt\)

\(v(t)=-\frac{A}{2}te^{-2t}-\frac{A}{4}e^{-2t}+C\)

Choose \(A=-2\) and \(C=0\), thus \(v(t)=te^{-2t}+\frac{1}{2}e^{-2t}\)

Conclusion:

Then, the function g becomes

\(g(t)=3(te^{-2t}+\frac{1}{2}e^{-2t})e^{2t}\)

\(=3t+\frac{3}{2}\)

Therefore, the second linearly independent solution is \(g(t)=3t+\frac{3}{2}\)

Obtain derivatives:

\(g'(t)=6ve^{2t}+3e^{2t}v'\)

and

\(g''(t)=12ve^{2t}+6e^{2t}v'+6e^{2t}v'+3e^{2t}v''\)

\(=12ve^{2t}+12e^{2t}v'+3e^{2t}v''\)

Substitute g and its derivative:

\(t(12ve^{2t}+12e^{2t}v'+3e^{2t}v'')-(2t+1)(6ve^{2t}+3e^{2t}v')+2(3ve^{2t})=0\)

\(12tve^{2t}+12te^{2t}v'+3te^{2t}v''-12tve^{2t}-6te^{2t}v'-6ve^{2t}-3e^{2t}v'+6ve^{2t}=0\)

\(3te^{2t}v''+(6t-3)e^{2t}v'=0\)

\(3tv''+(6t-3)v'=0\)

Further follows,

Let \(w=v'\), then

\(3tw'+(6t-3)w=0\)

\(3tw'=-(6t-3)w\)

\(\frac{dw}{w}=\frac{3-6t}{3t}dt\)

\(\frac{dw}{w}=(\frac{1}{t}-2)dt\)

Integrate both sides:

\(\int\frac{dw}{w}=\int(\frac{1}{2}-2)dt\)

\(\ln w=\ln t-2t+C\)

\(w=e^{\ln t-2t+C}=Ate^{-2t}\)

Then, we have

As \(w=v'\), obtain the function v(t):

\(v'=Ate^{-2t}\)

\(v(t)=-\frac{A}{2}te^{-2t}+\int\frac{A}{2}e^{-2t}dt\)

\(v(t)=-\frac{A}{2}te^{-2t}-\frac{A}{4}e^{-2t}+C\)

Choose \(A=-2\) and \(C=0\), thus \(v(t)=te^{-2t}+\frac{1}{2}e^{-2t}\)

Conclusion:

Then, the function g becomes

\(g(t)=3(te^{-2t}+\frac{1}{2}e^{-2t})e^{2t}\)

\(=3t+\frac{3}{2}\)

Therefore, the second linearly independent solution is \(g(t)=3t+\frac{3}{2}\)