Step 1

Divide equation with \(\displaystyle{\left({t}-{3}\right)}\),

\(\displaystyle{y}'+{\frac{{{\ln{{t}}}}}{{{t}-{3}}}}\ {y}={\frac{{{2}{t}}}{{{t}-{3}}}}\)

According to Theorem 2.4.1,

\(\displaystyle{p}{\left({t}\right)}={\frac{{{\ln{{t}}}}}{{{t}-{3}}}}\) and \(\displaystyle{g{{\left({t}\right)}}}={\frac{{{2}{t}}}{{{t}-{3}}}}\)

Functions p(t), g(t) are continuos on intervals (0, 3) and \(\displaystyle{\left({3},\ \infty\right)}\) because domains are \(\displaystyle{\frac{{{\left({0},\ \infty\right)}}}{{{\left\lbrace{3}\right\rbrace}}}}\) for p and \(\displaystyle{\frac{{{\mathbb{{{R}}}}{\left\lbrace{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{}}\) for g.

Since \(\displaystyle{t}_{{{0}}}={1}\), solution exists on (0, 3)

Divide equation with \(\displaystyle{\left({t}-{3}\right)}\),

\(\displaystyle{y}'+{\frac{{{\ln{{t}}}}}{{{t}-{3}}}}\ {y}={\frac{{{2}{t}}}{{{t}-{3}}}}\)

According to Theorem 2.4.1,

\(\displaystyle{p}{\left({t}\right)}={\frac{{{\ln{{t}}}}}{{{t}-{3}}}}\) and \(\displaystyle{g{{\left({t}\right)}}}={\frac{{{2}{t}}}{{{t}-{3}}}}\)

Functions p(t), g(t) are continuos on intervals (0, 3) and \(\displaystyle{\left({3},\ \infty\right)}\) because domains are \(\displaystyle{\frac{{{\left({0},\ \infty\right)}}}{{{\left\lbrace{3}\right\rbrace}}}}\) for p and \(\displaystyle{\frac{{{\mathbb{{{R}}}}{\left\lbrace{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{}}\) for g.

Since \(\displaystyle{t}_{{{0}}}={1}\), solution exists on (0, 3)