The equation provided is:

$t^2 y'' - t y' + y = 0$

We are informed that a solution exists in the form $y = t^m$, where $m \in \mathbb{N}$, and the task is to determine the general solution.

Steps to Solve:

1. Assume $y = t^m$, and calculate the derivatives: $y' = m t^{m-1}, \quad y'' = m(m-1)t^{m-2}.$

2. Substitute $y$, $y'$, and $y''$ into the differential equation: $t^2 \cdot m(m-1)t^{m-2} - t \cdot m t^{m-1} + t^m = 0.$

3. Simplify each term:

   • $t^2 \cdot m(m-1)t^{m-2} = m(m-1)t^m$,
   • $-t \cdot m t^{m-1} = -m t^m$,
   • $t^m = t^m$.

   Substituting these gives: $m(m-1)t^m - m t^m + t^m = 0.$

4. Factor out $t^m$ (since $t^m \neq 0$): $t^m \left[m(m-1) - m + 1\right] = 0.$

5. Simplify the expression inside the brackets: $m(m-1) - m + 1 = m^2 - m - m + 1 = m^2 - 2m + 1.$

6. Factorize: $m^2 - 2m + 1 = (m-1)^2.$

Thus: $t^m \cdot (m-1)^2 = 0.$

7. For the equation to hold, $(m-1)^2 = 0$, which implies: $m = 1.$

Particular Solution:

The particular solution is: $y = t^1 = t.$

General Solution:

Since the differential equation is linear and of second order, the general solution is a linear combination of independent solutions. The second independent solution is typically obtained by solving using methods like reduction of order. Here, the general solution is: $y(t) = C_1 t + C_2 t \ln(t),$ where $C_1$ and $C_2$ are arbitrary constants.

Let me know if you'd like a deeper explanation of any step! 😊

Related Questions:

1. What are the conditions for a differential equation to admit solutions of the form $y = t^m$?
2. How is the method of reduction of order applied to second-order differential equations?
3. What role do initial conditions play in determining the constants $C_1$ and $C_2$?
4. Can this method be extended to find solutions for equations of higher order?
5. How does the natural logarithmic term arise in the solution of linear differential equations?

Tip:

When testing solutions of the form $y = t^m$, always ensure that the parameter $m$ satisfies the resulting characteristic equation.