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

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Most popular questions from this chapter

A spring is stretched \(10 \mathrm{cm}\) by a force of 3 newtons. A mass of \(2 \mathrm{kg}\) is hung from the spring and is also attached to a viscous damper that exerts a force of 3 newtons when the velocity of the mass is \(5 \mathrm{m} / \mathrm{sec}\). If the mass is pulled down \(5 \mathrm{cm}\) below its equilibrium position and given an initial downward velocity of \(10 \mathrm{cm} / \mathrm{sec},\) determine its position \(u\) at any time \(t\) Find the quasi frequency \(\mu\) and the ratio of \(\mu\) to the natural frequency of the corresponding undamped motion.

Show that the period of motion of an undamped vibration of a mass hanging from a vertical spring is \(2 \pi \sqrt{L / g},\) where \(L\) is the elongation of the spring due to the mass and \(g\) is the acceleration due to gravity.

Verify that the given functions \(y_{1}\) and \(y_{2}\) satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. In Problems 19 and \(20 g\) is an arbitrary continuous function. $$ t y^{\prime \prime}-(1+t) y^{\prime}+y=t^{2} e^{2 i}, \quad t>0 ; \quad y_{1}(t)=1+t, \quad y_{2}(t)=e^{t} $$

Use the method of variation of parameters to find a particular solution of the given differential equation. Then check your answer by using the method of undetermined coefficients. $$ v^{\prime \prime}-v^{\prime}-2 v=2 e^{-t} $$

If \(a, b,\) and \(c\) are positive constants, show that all solutions of \(a y^{\prime \prime}+b y^{\prime}+c y=0\) approach zero as \(t \rightarrow \infty\).

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