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Transform the given equation into a system of first order equations. \(t^{2} u^{\prime \prime}+t u^{\prime}+\left(t^{2}-0.25\right) u=0\)

Short Answer

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Question: Transform the given second-order differential equation into a system of first-order equations: \(t^2 u''(t) + tu'(t) + (t^2 - 0.25)u(t) = 0\). Answer: The given second-order differential equation can be transformed into a system of two first-order equations as follows: (1) \(u'(t) = v(t)\), (2) \(t^2 v'(t) + t v(t) + (t^2 - 0.25)u(t) = 0\).

Step by step solution

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01

Introduce a new dependent variable for the first derivative of u

Let \(v(t) = u'(t)\). This new function represents the first derivative of \(u\).
02

Write the given equation in terms of the new variable

Substitute \(v\) for \(u'(t)\) and \(v'(t)\) for \(u''(t)\) in the given equation: \(t^2 v'(t) + t v(t) + (t^2 - 0.25)u(t) = 0\).
03

Separate the equation into a system of first-order equations

Write the original equation as two first-order equations with respect to \(u\) and \(v\): (1) \(u'(t) = v(t)\), (2) \(t^2 v'(t) + t v(t) + (t^2 - 0.25)u(t) = 0\), which represents a system of two first-order differential equations involving \(u(t)\) and \(v(t)\).

Key Concepts

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

Differential Equations
A differential equation is a mathematical equation that relates a function with its derivatives. These equations often model dynamic processes in physics, engineering, biology, and finance, among other fields. For example, they can describe how a physical system changes over time or space. The key to understanding a differential equation is knowing that the solution is not a single number, but a function or a set of functions that satisfy the given relationship between the variable and its derivative.

In the context of our exercise, the given equation involves a second derivative, which makes it a second-order differential equation. It's important for students to recognize that a higher-order differential equation can be transformed into a system of first-order differential equations, making the system easier to solve with standard methods.
Boundary Value Problem
Boundary value problems are a type of differential equation problem where we are given certain conditions, known as 'boundary conditions', that the solution must satisfy at the boundaries of the domain. These domains could be spatial boundaries for physical problems or time intervals for temporal problems.

While boundary conditions were not specified in our exercise, it's useful to note that solving most real-world problems requires taking these conditions into account. Boundary conditions help us determine the specific solution from the infinite family of solutions that differential equations usually have. In a classroom or textbook scenario, understanding how to apply boundary conditions is essential for correctly solving these types of problems.
Transformation of Equations
Sometimes, solving a differential equation directly can be very challenging or practically impossible. To cope with this, we use a strategy known as the transformation of equations. This involves manipulations that convert complex equations into simpler forms or systems of equations that are more tractable.

In our exercise, the transformation involved introducing a new variable to convert the second-order differential equation into a system of first-order differential equations. This step simplifies the problem and opens up the possibility of using methods designed specifically for first-order systems, such as the method of characteristics or matrix exponentials for linear systems. Transformation techniques are invaluable tools in a mathematician's arsenal.
Ordinary Differential Equations
Ordinary differential equations (ODEs) are differential equations that contain one or more functions of one independent variable and its derivatives. They differ from partial differential equations (PDEs), which involve multiple independent variables. ODEs are termed 'ordinary' to indicate that they have a single variable, in contrast to the 'partial', which indicates multiple variables in PDEs.

The system we dealt with in this exercise consists of ODEs, as there is only one independent variable, which is time (t). Understanding the classification between ODEs and PDEs is crucial when selecting appropriate solution methods. For example, methods for solving first-order ODEs include separation of variables, integrating factors, and linear algebra techniques for systems of linear ODEs.

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Most popular questions from this chapter

In each of Problems 13 through 20 the coefficient matrix contains a parameter \(\alpha\). In each of these problems: (a) Determine the eigervalues in terms of \(\alpha\). (b) Find the critical value or values of \(\alpha\) where the qualitative nature of the phase portrait for the system changes. (c) Draw a phase portrait for a value of \(\alpha\) slightly below, and for another value slightly above, each crititical value. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{\alpha} & {1} \\ {-1} & {\alpha}\end{array}\right) \mathbf{x} $$

Deal with the problem of solving \(\mathbf{A x}=\mathbf{b}\) when \(\operatorname{det} \mathbf{A}=0\) Suppose that det \(\mathbf{\Lambda}=0,\) and that \(\mathbf{x}=\mathbf{x}^{(0)}\) is a solution of \(\mathbf{A} \mathbf{x}=\mathbf{b} .\) Show that if \(\xi\) is a solution of \(\mathbf{A} \xi=\mathbf{0}\) and \(\alpha\) is any constant, then \(\mathbf{x}=\mathbf{x}^{(0)}+\alpha \xi\) is also a solution of \(\mathbf{A} \mathbf{x}=\mathbf{b} .\)

Let \(\mathbf{J}=\left(\begin{array}{cc}{\lambda} & {1} \\ {0} & {\lambda}\end{array}\right),\) where \(\lambda\) is an arbitrary real number. (a) Find \(\mathbf{J}^{2}, \mathbf{J}^{3},\) and \(\mathbf{J}^{4}\) (b) Use an inductive argument to show that \(\mathbf{J}^{n}=\left(\begin{array}{cc}{\lambda^{n}} & {n \lambda^{n-1}} \\\ {0} & {\lambda^{n}}\end{array}\right)\) (c) Determine exp(Jt). (d) Use exp(Jt) to solve the initial value problem \(\mathbf{x}^{\prime}=\mathbf{J x}, \mathbf{x}(0)=\mathbf{x}^{0}\)

Consider the initial value problem $$ x^{\prime}=A x+g(t), \quad x(0)=x^{0} $$ (a) By referring to Problem \(15(c)\) in Section \(7.7,\) show that $$ x=\Phi(t) x^{0}+\int_{0}^{t} \Phi(t-s) g(s) d s $$ (b) Show also that $$ x=\exp (A t) x^{0}+\int_{0}^{t} \exp [\mathbf{A}(t-s)] \mathbf{g}(s) d s $$ Compare these results with those of Problem 27 in Section \(3.7 .\)

(a) Find the eigenvalues of the given system. (b) Choose an initial point (other than the origin) and draw the corresponding trajectory in the \(x_{1} x_{2}\) -plane. Also draw the trajectories in the \(x_{1} x_{1}-\) and \(x_{2} x_{3}-\) planes. (c) For the initial point in part (b) draw the corresponding trajectory in \(x_{1} x_{2} x_{3}\) -space. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rrr}{-\frac{1}{4}} & {1} & {0} \\\ {-1} & {-\frac{1}{4}} & {0} \\ {0} & {0} & {\frac{1}{10}}\end{array}\right) \mathbf{x} $$

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