solve t^3y'''-t^2y''+2ty'-2y = 0 with y(1) =3, y'(1)=2

We are given a third-order linear differential equation with variable coefficient to solve. The equation is:
$t^3{y^{″}}^{\prime }-t^2{y}^{″}+2t{y}^{\prime }-2y=0$
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:
$y^{\prime }(t)={\sum }_{n=1}^{\infty }n{a}_n{t}^{n-1}$
$y^{″}(t)={\sum }_{n=2}^{\infty }n(n-1)a_n{t}^{n-2}$
${y^{″}}^{\prime }(t)={\sum }_{n=3}^{\infty }n(n-1)(n-2)a_n{t}^{n-3}$
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:
${\sum }_{n=0}^{\infty }{a}_n{t}^{n+2}{n}^3+3a_n{t}^{n+2}{n}^2=0$
Equating the coefficients of like powers of t to zero, we get the following recurrence relation:
$a_{n+3}=\frac{3a_{n+2}}{(n+2)(n+1)}\text{for}n\ge 0$
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:
$c^3-3c^2+2c=0$
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:
$y(1)=a_0+a_1+a_2+\dots =a_0+{\sum }_{n=0}^{\infty }{a}_{n+1}=a_0+{\sum }_{n=1}^{\infty }{a}_n{t}^n=3$
$y^{\prime }(1)=a_1+2a_2+3a_3+\dots ={\sum }_{n=0}^{\infty }(n+1)a_{n+1}=2$
$y^{″}(1)=2a_2+6a_3+12a_4+\dots ={\sum }_{n=0}^{\infty }(n+2)(n+1)a_{n+2}=1$
Substituting the expressions for $a_n$ and simplifying, we get:
$a_0+\frac{3}{2}{a}_3+\frac{15}{4}{a}_4+\dots =3$
$a_1+4a_2+\frac{15}{2}{a}_3+\dots =2$
$2a_2+12a_3+36a_4+\dots =1$
Using the first equation to eliminate $a_0$, and the third equation to eliminate $a_4$, we get:
$\frac{9}{4}{a}_3+\dots =3-\frac{3}{2}{a}_3$
$2a_2+12a_3+\dots =1$
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:
$a_n=\frac{3(-1{)}^n}{4·2^n}\left(\genfrac{}{}{0ex}{}{n+2}{2}\right)-\frac{7(-1{)}^n}{24}\left(\genfrac{}{}{0ex}{}{n+1}{1}\right)+\frac{32}{81}{\delta }_{n3}$
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 $y^{″}(1)=1$ is:
$\boxed{y(t)=3-\frac{7}{24}{t}^2+\frac{32}{81}{t}^3+\dots }$