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How many arbitrary constants appear in the general solution of \(y^{\prime \prime}(t)+9 y(t)=10 ?\)

Short Answer

Expert verified
Answer: There are 2 arbitrary constants in the general solution of the given ODE.

Step by step solution

01

Rewrite the ODE

Rewrite the given second-order linear ODE as \(y^{\prime\prime}(t) = -9y(t) + 10\).
02

Solve the corresponding homogeneous ODE

Solve the homogeneous ODE \(y^{\prime\prime}(t) + 9y(t) = 0\). To do this, we will first consider its characteristic equation: \(λ^2 + 9 = 0\) \(Learners will solve for the eigenvalues \)λ\( to get \)λ=\pm 3i$
03

Find the general solution of the homogeneous ODE

Using the eigenvalues \(\pm 3i\) from Step 2, we can find the general solution for the homogeneous ODE. The general solution will be of the form: y(t) = \(C_1 \cdot \sin(3t) + C_2 \cdot \cos(3t)\) where \(C_1, C_2\) are arbitrary constants.
04

Formulate the general solution of the given non-homogeneous ODE

Since the given ODE is non-homogeneous, we also need to find a particular solution, which can be done using various methods like undetermined coefficients, variation of parameters, etc. Once the particular solution is found, the general solution of the given non-homogeneous ODE will be the sum of the general solution of the homogeneous ODE and the particular solution. Let's denote the particular solution as \(y_p(t)\). The general solution of the given ODE: y(t) = \(C_1 \cdot \sin(3t) + C_2 \cdot \cos(3t) + y_p(t)\)
05

Count the arbitrary constants

Looking at the general solution found in Step 4, we see two arbitrary constants: \(C_1\) and \(C_2\). Thus, the answer to the question is that there are 2 arbitrary constants in the general solution of the given ODE.

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Key Concepts

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

Second-order ODE
In mathematical terms, a second-order ordinary differential equation (ODE) refers to an equation that involves the second derivative of a function. This means it models a scenario where the rate of change of a rate of change is of interest. For instance, it could model physical phenomena such as the motion of a pendulum or the oscillation of a spring. Second-order ODEs usually take the form:\[y''(t) + f(t)y'(t) + g(t)y(t) = h(t)\]where \(y''(t)\) is the second derivative, \(y'(t)\) is the first derivative, and \(y(t)\) is the function itself. These equations can be linear or nonlinear, but linear equations such as the one given in the problem are more common in applications. When solving second-order ODEs, it is crucial to specify initial conditions or boundary conditions to pin down a unique solution among the infinite suite of possible solutions. Moreover, these ODEs can be homogeneous or non-homogeneous, each presenting unique methods for finding a solution.
Homogeneous Equations
A homogeneous differential equation is characterized by the fact that the entire function on the right-hand side of the equation is zero, that is:\[y''(t) + 9y(t) = 0\]For such equations, all terms are dependent on the function and its derivatives, without any forcing function (i.e., an external influence). This means homogeneous equations model naturally occurring behavior without an external influence. To solve these equations, one generally derives a characteristic equation, which is a polynomial obtained from substituting a trial solution (usually exponential functions) into the ODE. For example, solving our characteristic equation:\[λ^2 + 9 = 0\]leads to complex eigenvalues \(λ = \pm 3i\). These values suggest that the solution will have oscillatory components due to the imaginary parts.
  • The solution form \(C_1 \cdot \sin(3t) + C_2 \cdot \cos(3t)\) demonstrates the typical sine and cosine functions that emerge from solutions to homogeneous equations with complex eigenvalues.
The constants \(C_1\) and \(C_2\) are arbitrary constants, which are determined by initial or boundary conditions.
Particular Solution
In contrast to homogeneous equations, non-homogeneous ODEs involve an additional term that represents an external influence or "forcing function." These ODEs take the general form:\[y''(t) + 9y(t) = 10\]where the constant term (in this case, 10) is the non-homogeneous part that adds complexity. To solve for the particular solution, which specifically accounts for this influence, techniques like undetermined coefficients or variation of parameters are employed.The particular solution is essentially a solution to the non-homogeneous equation not considering initial conditions, but it complements the homogeneous solution to account for the entire behavior described by the ODE. Adding this solution to the general solution of the homogeneous equation:\[y(t) = C_1 \cdot \sin(3t) + C_2 \cdot \cos(3t) + y_p(t)\]completes the general solution of the non-homogeneous ODE. Here
  • \(y_p(t)\) is a specific solution that satisfies the non-homogeneous equation.
The sum integrates both the natural behavior modeled by the homogeneous part and the forced response described by the non-homogeneous component.

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Most popular questions from this chapter

Use a calculator or computer program to carry out the following steps. a. Approximate the value of \(y(T)\) using Euler's method with the given time step on the interval \([0, T]\). b. Using the exact solution (also given), find the error in the approximation to \(y(T)\) (only at the right endpoint of the time interval). c. Repeating parts (a) and (b) using half the time step used in those calculations, again find an approximation to \(y(T)\). d. Compare the errors in the approximations to \(y(T)\). $$y^{\prime}(t)=-2 y, y(0)=1 ; \Delta t=0.2, T=2 ; y(t)=e^{-2 t}$$

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Solve the following initial value problems. When possible, give the solution as an explicit function of \(t\) $$y^{\prime}(t)=\frac{\cos ^{2} t}{2 y}, y(0)=-2$$

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RC circuit equation Suppose a battery with voltage \(V\) is connected in series to a capacitor (a charge storage device) with capacitance \(C\) and a resistor with resistance \(R\). As the charge \(Q\) in the capacitor increases, the current \(I\) across the capacitor decreases according to the following initial value problems. Solve each initial value problem and interpret the solution. a. \(I^{\prime}(t)+\frac{1}{R C} I(t)=0, I(0)=\frac{V}{R}\) b. \(Q^{\prime}(t)+\frac{1}{R C} Q(t)=\frac{V}{R}, Q(0)=0\)

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