Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The method of Problem 20 can be extended to second order equations with variable coefficients. If \(y_{1}\) is a known nonvanishing solution of \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0,\) show that a second solution \(y_{2}\) satisfies \(\left(y_{2} / y_{1}\right)^{\prime}=W\left(y_{1}, y_{2}\right) / y_{1}^{2},\) where \(W\left(y_{1}, y_{2}\right)\) is the Wronskian \(\left. \text { of }\left.y_{1} \text { and } y_{2} \text { . Then use Abel's formula [Eq. ( } 8\right) \text { of Section } 3.3\right]\) to determine \(y_{2}\).

Short Answer

Expert verified
Question: Show that if \(y_1\) is a known nonvanishing solution of the second-order linear homogeneous differential equation \(y'' + p(t)y' + q(t)y = 0\), then another solution \(y_2\) satisfies the equation \(\frac{d\left(y_2/y_1\right)}{dt}=\frac{W(y_1, y_2)}{y_1^2}\), where \(W(y_1,y_2)\) is the Wronskian of \(y_1\) and \(y_2\). Solution: We can show that \(\frac{d\left(y_2/y_1\right)}{dt}=\frac{W(y_1,y_2)}{y_1^2}\) by following these steps: 1. Compute the Wronskian of \(y_1\) and \(y_2\): \(W(y_1,y_2)=y_1y_2' - y_2y_1'\) 2. Show that \(\frac{d\left(y_2/y_1\right)}{dt}=\frac{W(y_1, y_2)}{y_1^2}\): We find that \(\frac{d\, (y_2/y_1)}{dt} = \frac{y_1 y_2' - y_2 y_1'}{y_1^2}\), which is the same as \(W(y_1, y_2)/y_1^2\) 3. Use Abel's formula: Abel's formula states that the Wronskian of two solutions of a homogeneous linear differential equation is given by \(W(y_1,y_2)=C e^{-\int p(t)\,dt}\), where \(C\) is a constant. 4. Determine \(y_2\): Integrating the expression \(\frac{d\, (y_2/y_1)}{dt} = Ce^{-\int p(t)\,dt}\) and solving for \(y_2\), we find the second solution: \(y_2 = y_1 \left(\int Ce^{-\int p(t)\,dt} dt + C_1\right)\), where \(C_1\) is another constant of integration.

Step by step solution

01

Compute Wronskian of \(y_1\) and \(y_2\)

To compute the Wronskian \(W(y_1,y_2)\), we need to evaluate the determinant of a matrix formed by the functions and their first derivatives: $$W(y_1,y_2)= \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} =y_1y_2' - y_2y_1'$$
02

Show that \(\frac{d\left(y_2/y_1\right)}{dt}=\frac{W(y_1, y_2)}{y_1^2}\)

To show that the expression is true, we first compute the derivative of \(y_2/y_1\) with respect to \(t\): $$\frac{d\, (y_2/y_1)}{dt} = \frac{y_1 y_2' - y_2 y_1'}{y_1^2}$$ Comparing this expression with the Wronskian, we see that they are equal: $$\frac{d\, (y_2/y_1)}{dt} = \frac{W(y_1, y_2)}{y_1^2}$$
03

Use Abel's Formula

Abel's formula states that the Wronskian of two solutions of a homogeneous linear differential equation is given by: $$W(y_1,y_2)=C e^{-\int p(t)\,dt}$$ where \(C\) is a constant. Therefore, we can write $$\frac{W(y_1, y_2)}{y_1^2}=Ce^{-\int p(t)\,dt}$$ Combining this with the expression found in Step 2 leads to: $$\frac{d\, (y_2/y_1)}{dt} = Ce^{-\int p(t)\,dt}$$
04

Determine \(y_2\)

To find \(y_2\), we integrate the above expression: $$\int \frac{d\, (y_2/y_1)}{dt} dt = \int Ce^{-\int p(t)\,dt} dt$$ The left-hand side of the equation represents the integral of a derivative, so we have $$y_2/y_1 = \int Ce^{-\int p(t)\,dt} dt + C_1$$ where \(C_1\) is another constant of integration. Finally, we can solve for \(y_2\): $$y_2 = y_1 \left(\int Ce^{-\int p(t)\,dt} dt + C_1\right)$$ This expression gives us the second solution \(y_2\) in terms of the first solution \(y_1\) and the coefficients of the differential equation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Introduction to Second Order Linear Differential Equations
Second order linear differential equations play a significant role in various fields, including physics, engineering, and economics. Understanding their nature and how to solve them is crucial for many scientific and technical applications. The central challenge lies in finding the general solution, which often involves identifying two linearly independent solutions. When dealing with variable coefficients and no straightforward methods apply, it is essential to leverage specialized techniques.
Understanding the Wronskian
The Wronskian is a determinant used in the analysis of differential equations to determine if a set of functions are linearly independent. In the context of second order linear differential equations with variable coefficients, if you have two solutions, say y1 and y2, their Wronskian is calculated as:

\[ W(y_1,y_2) = y_1y_2' - y_2y_1' \]
The Wronskian is not only a test for linear independence but also plays a critical role in finding a second solution to a homogenous equation once one solution is already known.
Variable Coefficients and Their Implications
Differential equations with variable coefficients, expressed as functions p(t) and q(t) in the equation \(y'' + p(t)y' + q(t)y = 0\), are often more complex than those with constant coefficients because the variable coefficients make the equation non-uniform. This leads to solutions that are not straightforward and typically do not have a set form. Solving these requires a more in-depth analysis or the use of special methods like the method of variation of parameters.
Harnessing Abel's Formula
Abel's formula provides a way to compute the Wronskian without solving the entire differential equation, by relating it to the coefficient p(t) of the first derivative term in our homogeneous equation. Abel's formula states:

\[ W(y_1,y_2) = C e^{-\boldsymbol{\textstyle \text{∫}} p(t) \boldsymbol{\text{dt}}} \]
where C is a constant. This formula is extraordinarily useful when solving for a second linearly independent solution to the differential equation, as shown in the exercise solution.
Method of Variation of Parameters
The method of variation of parameters is a technique used when standard methods, such as the method of undetermined coefficients, cannot be applied to non-homogeneous equations or those with variable coefficients. This method involves using the solution of the associated homogeneous equation to find a particular solution for the non-homogeneous equation. The key to variation of parameters is the assumption that the constants in the homogeneous solution are indeed variable, allowing us to adjust them to fit the non-homogeneous equation, ultimately leading to a complete solution of the original differential equation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the general solution of the given differential equation. $$ 2 y^{\prime \prime}-3 y^{\prime}+y=0 $$

Use the method of Problem 32 to solve the given differential $$ 2 y^{\prime \prime}+3 y^{\prime}+y=t^{2}+3 \sin t \quad(\text { see Problem } 7) $$

By combining the results of Problems 24 through \(26,\) show that the solution of the initial value problem $$ L[y]=\left(a D^{2}+b D+c\right) y=g(t), \quad y\left(t_{0}\right)=0, \quad y^{\prime}\left(t_{0}\right)=0 $$ where \(a, b,\) and \(c\) are constants, has the form $$ y=\phi(t)=\int_{t_{0}}^{t} K(t-s) g(s) d s $$ The function \(K\) depends only on the solutions \(y_{1}\) and \(y_{2}\) of the corresponding homogeneous equation and is independent of the nonhomogeneous term. Once \(K\) is determined, all nonhomogeneous problems involving the same differential operator \(L\) are reduced to the evaluation of an integral. Note also that although \(K\) depends on both \(t\) and \(s,\) only the combination \(t-s\) appears, so \(K\) is actually a function of a single variable. Thinking of \(g(t)\) as the input to the problem and \(\phi(t)\) as the output, it follows from Eq. (i) that the output depends on the input over the entire interval from the initial point \(t_{0}\) to the current value \(t .\) The integral in Eq. (i) is called the convolution of \(K\) and \(g,\) and \(K\) is referred to as the kernel.

By choosing the lower limitofation in Eq. ( 28 ) inthe textas the initial point \(t_{0}\), show that \(Y(t)\) becomes $$ Y(t)=\int_{t_{0}}^{t} \frac{y_{1}(s) y_{2}(t)-y_{1}(s) y_{2}(s)}{y_{1}(s) y_{2}^{2}(s)-y_{1}^{\prime}(s) y_{2}(s)} g(s) d s $$ Show that \(Y(t)\) is asolution of the initial value problem $$ L[y], \quad y\left(t_{0}\right)=0, \quad y^{\prime}\left(t_{0}\right)=0 $$ Thus \(Y\) can be identific d writh \(v\) in Problem \(21 .\)

Use the method of Problem 33 to find a second independent solution of the given equation. \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=0, \quad x>0 ; \quad y_{1}(x)=x^{-1 / 2} \sin x\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free