Question

Use the method of Problem 33 to find a second independent solution of the given equation. \(t^{2} y^{\prime \prime}+3 t y^{\prime}+y=0, \quad t>0 ; \quad y_{1}(t)=t^{-1}\)

Short answer

Answer: The second independent solution to the given differential equation is \(y_2(t) = \frac{3}{2t^3}\).

Step-by-step solution

(1. Write down the given differential equation and first solution)

The given differential equation and first solution are \(t^2\cdot y''+3t\cdot y'+y=0\), where \(y_1(t)=t^{-1}\).

(2. Wronskian of the two solutions)

The Wronskian, \(W(y_1, y_2)\), of the two solutions \(\frac{1}{t}\) and \(y_2(t)\) can be written as: \[W(y_1, y_2) = \begin{vmatrix} y_1 & y_2\\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} \frac{1}{t} & y_2 \\ -\frac{1}{t^2} & y_2' \end{vmatrix}\]

(3. Apply the method of Problem 33)

According to the method of Problem 33, we can find the second independent solution, \(y_2(t)\), by finding a function \(v(t)\) and formulating \(y_2(t)=v(t)y_1(t)\). We are given that \(\frac{1}{t}\) is the first solution to the given differential equation.

(4. Use Abel's Formula to find v(t))

The Abel's Formula states: \[W(y_1, y_2) = y_1y_2' - y_1'y_2 = \frac{1}{t} y_2' + \frac{1}{t^2} y_2 = c \cdot e^{-\int \frac{p(t)}{t} dt}\] where \(p(t)=\frac{3t}{t^2}\) in this case, which makes the integral \(-3 \ln(t) + C\). The Wronskian is a constant in this case, so we can assume \(c=1\) without loss of generality and exploit Abel's formula to find v(t) as follows: \[v'(t) = -\frac{3}{t^3} \implies v(t) = \int \left(-\frac{3}{t^3} \right) dt = \frac{3}{2t^2} + C\] Now, assuming \(C = 0\) (since we are looking for a solution linearly independent of the existing solution), we have \(v(t) = \frac{3}{2t^2}\).

(5. Find the second independent solution y_2(t))

The second independent solution can be found by multiplying \(v(t)\) by \(y_1(t)\). \[y_2(t) = v(t) \cdot y_1(t) = \left(\frac{3}{2t^2}\right) \cdot \frac{1}{t} = \boxed{\frac{3}{2t^3}}\] Thus, the second independent solution to the given differential equation is \(y_2(t) = \frac{3}{2t^3}\).

Key concepts

Wronskian

The Wronskian is a determinant used to assess whether a set of solutions is linearly independent. It is particularly important in solving systems of linear differential equations. For two functions, the Wronskian is represented as:

• \(W(y_1, y_2) = \begin{vmatrix} y_1 & y_2\ y_1' & y_2' \end{vmatrix}\).

In simpler terms, it checks if one function is a scalar multiple of another. If the Wronskian is not zero, the functions are linearly independent. For instance, in our original task:- The Wronskian for \( \frac{1}{t} \) and \( y_2(t) \) is the determinant of the matrix consisting of these functions and their derivatives. This determinant helps us explore whether \( y_2(t) \) truly offers a new solution different from \( \frac{1}{t} \).

Abel's formula

Abel's formula is a powerful tool in the realm of differential equations, simplifying the process of finding a Wronskian. Specifically, it aids in determining the constancy of the Wronskian for linear differential equations of the form \( t^2y'' + 3ty' + y = 0 \).Abel's formula states:

• \[W(y_1, y_2) = c \cdot e^{\int -p(t)/t \cdot dt}\]

Here, \( p(t) \) is obtained from the coefficient of \( y' \) in the differential equation. This expression simplifies this problem, showing that the Wronskian is either zero or a constant. Knowing that the Wronskian is constant, you can use it to find a function \( v(t) \) for the second solution's formula. With the integral equaling \(-3 \ln(t) \), it informs us about the nature of \( v(t) \), guiding us to achieve:

• \( v(t) = \frac{3}{2t^2} \), making sure \( y_2(t) = v(t) \times y_1(t) \).

linearly independent solutions

Linearly independent solutions are crucial when solving differential equations. They ensure diverse and non-redundant solutions, representing different directions or dimensions of the solution space.A set of solutions \( \{ y_1, y_2, \ldots, y_n \} \) is linearly independent if no solution can be expressed as a linear combination of others. The implication is profound:

• If combined appropriately, they can form a general solution for the differential equation.

In our context, we start with a known solution \( y_1(t) = \frac{1}{t} \). To ensure \( y_2(t) \) is independent, it must offer something new—thus the check via the Wronskian.Using the derivations:

• \( y_2(t) = \frac{3}{2t^3} \), multiplying the function \( v(t) \) by \( y_1(t) \), verifying distinctness.

This approach guarantees we have an appropriately independent solution in different behavioral states.