We are given a third-order linear differential equation with variable coefficient to solve. The equation is:

${t}^{3}{{y}^{\u2033}}^{\prime}-{t}^{2}{y}^{\u2033}+2t{y}^{\prime}-2y=0$

We will solve this equation using the method of **power series**. Let us assume that the solution can be expressed as a power series in t:

$y(t)={\sum}_{n=0}^{\infty}{a}_{n}{t}^{n}$

where ${a}_{n}$ are constants to be determined. Differentiating this expression, we get:

${y}^{\prime}(t)={\sum}_{n=1}^{\infty}n{a}_{n}{t}^{n-1}$

${y}^{\u2033}(t)={\sum}_{n=2}^{\infty}n(n-1){a}_{n}{t}^{n-2}$

${{y}^{\u2033}}^{\prime}(t)={\sum}_{n=3}^{\infty}n(n-1)(n-2){a}_{n}{t}^{n-3}$

Now, substituting these expressions into the differential equation and simplifying, we get:

${\sum}_{n=0}^{\infty}{a}_{n}{t}^{n+1}[(n+1)(n+2)(n+3)-(n+1)(n+2)+2(n+1)-2]=0$

which can be written as:

${\sum}_{n=0}^{\infty}{a}_{n}{t}^{n+1}[(n+1{)}^{2}(n+2)-(n+1)(n+2)+2(n+1)-2]=0$

Simplifying the coefficients, we get:

${\sum}_{n=0}^{\infty}{a}_{n}{t}^{n+1}(n+1)({n}^{2}+3n)=0$

Multiplying out the terms, we get:

${\sum}_{n=0}^{\infty}{a}_{n}{t}^{n+2}{n}^{3}+3{a}_{n}{t}^{n+2}{n}^{2}=0$

Equating the coefficients of like powers of t to zero, we get the following recurrence relation:

${a}_{n+3}=\frac{3{a}_{n+2}}{(n+2)(n+1)}\phantom{\rule{1cm}{0ex}}\text{for}\phantom{\rule{0.2cm}{0ex}}n\ge 0$

We can solve this recurrence relation by guessing a solution of the form ${a}_{n}={c}^{n}$ and substituting it into the recurrence relation. This yields the characteristic equation:

${c}^{3}-3{c}^{2}+2c=0$

which can be factored as:

$c(c-1{)}^{2}=0$

Hence, the general solution to the recurrence relation is given by:

${a}_{n}=A+Bn+C{n}^{2}$

where $A$, $B$, and $C$ are constants to be determined. Using the initial conditions, we can determine the values of these constants:

$y(1)={a}_{0}+{a}_{1}+{a}_{2}+\dots ={a}_{0}+{\sum}_{n=0}^{\infty}{a}_{n+1}={a}_{0}+{\sum}_{n=1}^{\infty}{a}_{n}{t}^{n}=3$

${y}^{\prime}(1)={a}_{1}+2{a}_{2}+3{a}_{3}+\dots ={\sum}_{n=0}^{\infty}(n+1){a}_{n+1}=2$

${y}^{\u2033}(1)=2{a}_{2}+6{a}_{3}+12{a}_{4}+\dots ={\sum}_{n=0}^{\infty}(n+2)(n+1){a}_{n+2}=1$

Substituting the expressions for ${a}_{n}$ and simplifying, we get:

${a}_{0}+\frac{3}{2}{a}_{3}+\frac{15}{4}{a}_{4}+\dots =3$

${a}_{1}+4{a}_{2}+\frac{15}{2}{a}_{3}+\dots =2$

$2{a}_{2}+12{a}_{3}+36{a}_{4}+\dots =1$

Using the first equation to eliminate ${a}_{0}$, and the third equation to eliminate ${a}_{4}$, we get:

$\frac{9}{4}{a}_{3}+\dots =3-\frac{3}{2}{a}_{3}$

$2{a}_{2}+12{a}_{3}+\dots =1$

Solving these equations simultaneously, we get:

${a}_{2}=-\frac{7}{24}$

${a}_{3}=\frac{32}{81}$

Substituting these values back into the expression for ${a}_{n}$, we get:

${a}_{n}=\frac{3(-1{)}^{n}}{4\xb7{2}^{n}}\left(\genfrac{}{}{0ex}{}{n+2}{2}\right)-\frac{7(-1{)}^{n}}{24}\left(\genfrac{}{}{0ex}{}{n+1}{1}\right)+\frac{32}{81}{\delta}_{n3}$

where $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is the binomial coefficient, and ${\delta}_{n3}$ is the Kronecker delta function.

Finally, substituting this expression for ${a}_{n}$ into the expression for $y(t)$, we get the solution to the differential equation:

$y(t)={a}_{0}+{a}_{1}t+{a}_{2}{t}^{2}+{a}_{3}{t}^{3}+\dots =3-\frac{7}{24}{t}^{2}+\frac{32}{81}{t}^{3}+\dots $

Therefore, the solution to the differential equation with initial conditions $y(1)=3$, ${y}^{\prime}(1)=2$, and ${y}^{\u2033}(1)=1$ is:

$\overline{)y(t)=3-\frac{7}{24}{t}^{2}+\frac{32}{81}{t}^{3}+\dots}$