Chapter 3: Problem 34
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
Expert verified
Answer: The second independent solution to the given differential equation is \(y_2(t) = \frac{3}{2t^3}\).
Step by step solution
01
(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 \(y_1(t)=t^{-1}\).
02
(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:
\[W(y_1, y_2) = \begin{vmatrix} y_1 & y_2\\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} \frac{1}{t} & y_2 \\ -\frac{1}{t^2} & y_2' \end{vmatrix}\]
03
(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.
04
(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 \(v(t) = \frac{3}{2t^2}\).
05
(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 \(y_2(t) = \frac{3}{2t^3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
- \(W(y_1, y_2) = \begin{vmatrix} y_1 & y_2\ y_1' & y_2' \end{vmatrix}\).
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:
- \[W(y_1, y_2) = c \cdot e^{\int -p(t)/t \cdot dt}\]
- \( v(t) = \frac{3}{2t^2} \), 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.
- \( y_2(t) = \frac{3}{2t^3} \), multiplying the function \( v(t) \) by \( y_1(t) \), verifying distinctness.