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Chapter 3: Problem 8

In each of Problems 1 through 10 find the general solution of the given differential equation. \(16 y^{\prime \prime}+24 y^{\prime}+9 y=0\)

Short Answer

Expert verified
Answer: The general solution of the given differential equation is \(y(x) = (C_1 + C_2 x) e^{-\frac{3}{4}x}\).

Step by step solution

01

Find the Characteristic Equation and Roots:

The characteristic equation corresponding to the differential equation \(16y^{\prime\prime} + 24y^{\prime} + 9y = 0\) is given by: \(16r^2 + 24r + 9 = 0\) We need to find the roots of this quadratic equation.
02

Use the quadratic formula:

We will use the quadratic formula to find the roots of the characteristic equation: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 16\), \(b = 24\), and \(c = 9\). \(r = \frac{-24 \pm \sqrt{24^2 - 4(16)(9)}}{2(16)}\) \(r = \frac{-24 \pm \sqrt{576 - 576}}{32}\) \(r = \frac{-24}{32}\) Since the discriminant in the root formula is zero, both roots r are equal: \(r_1 = r_2 = -\frac{3}{4}\)
03

Determine the general solution:

For a second-order linear homogeneous differential equation with constant coefficients when we have real and repeated roots, the general solution is given by: \(y(x) = (C_1 + C_2 x) e^{rx}\) Here, we found that \(r_1 = r_2 = -\frac{3}{4}\). So, the general solution for the given differential equation is: \(y(x) = (C_1 + C_2 x) e^{-\frac{3}{4}x}\) This general solution represents all possible solutions of the given differential equation.

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Key Concepts

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

Characteristic Equation
The characteristic equation is a vital tool in solving linear differential equations. It transforms the differential equation into an algebraic equation. This makes it easier to handle. In our example, the differential equation is:
\[16y^{\prime\prime} + 24y^{\prime} + 9y = 0\]
To find the characteristic equation, replace derivatives with powers of \(r\). This gives us:
\[16r^2 + 24r + 9 = 0\]
This captures the essence of the original differential equation. It's simpler to solve. The roots of this characteristic equation help determine the form of the solution to the differential equation.
Quadratic Formula
Once the characteristic equation is found, the next step is often to solve it using the quadratic formula. This formula helps find the roots of any quadratic equation of the form \(ax^2 + bx + c = 0\).
The formula is stated as:
\[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In our problem, the terms are:
  • \(a = 16\)
  • \(b = 24\)
  • \(c = 9\)
Plug these into the formula:
\[r = \frac{-24 \pm \sqrt{24^2 - 4(16)(9)}}{2(16)}\]
This results in \(r = -\frac{3}{4}\) as the roots. Notice the discriminant (the part under the square root) is zero. Thus, both roots are equal.
General Solution
The general solution of a differential equation like ours is derived from the roots of the characteristic equation. Because our roots were real and equal, the general solution takes a specific form:
\[y(x) = (C_1 + C_2 x) e^{rx}\]
Substitute the found root \(r = -\frac{3}{4}\) into this formula to get:
\[y(x) = (C_1 + C_2 x) e^{-\frac{3}{4}x}\]
This formula captures every possible solution for the differential equation. It includes constants \(C_1\) and \(C_2\), which can be adjusted based on initial conditions. This flexibility represents the infinite nature of solutions available for such equations.
Homogeneous Equation
In differential equation terms, a homogeneous equation implies all terms are dependent on the function and its derivatives. Our equation:
\[16y^{\prime\prime} + 24y^{\prime} + 9y = 0\]
is homogeneous because everything equals zero. There are no standalone constant terms.
Homogeneous equations often have standard forms that help simplify the solving process. This involves finding the characteristic equation as a first step.
  • Transform the equation to a polynomial.
  • Solve using methods like the quadratic formula.
  • Find a general solution based on the roots.
This makes them easier to manage. It's due to the predictable nature of their solutions.

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Most popular questions from this chapter

Deal with the initial value problem $$ u^{\prime \prime}+0.125 u^{\prime}+u=F(t), \quad u(0)=2, \quad u^{\prime}(0)=0 $$ (a) Plot the given forcing function \(F(t)\) versus \(t\) and also plot the solution \(u(t)\) versus \(t\) on the same set of axes. Use a \(t\) interval that is long enough so the initial transients are substantially eliminated. Observe the relation between the amplitude and phase of the forcing term and the amplitude and phase of the response. Note that \(\omega_{0}=\sqrt{k / m}=1\). (b) Draw the phase plot of the solution, that is, plot \(u^{\prime}\) versus \(u .\) \(F(t)=3 \cos (0.3 t)\)

If a series circuit has a capacitor of \(C=0.8 \times 10^{-6}\) farad and an inductor of \(L=0.2\) henry, find the resistance \(R\) so that the circuit is critically damped.

A mass of \(20 \mathrm{g}\) stretches a spring \(5 \mathrm{cm}\). Suppose that the mass is also attached to a viscous damper with a damping constant of \(400 \mathrm{dyne}\) -sec/cm. If the mass is pulled down an additional \(2 \mathrm{cm}\) and then released, find its position \(u\) at any time \(t .\) Plot \(u\) versus \(t .\) Determine the quasi frequency and the quasi period. Determine the ratio of the quasi period to the period of the corresponding undamped motion. Also find the time \(\tau\) such that \(|u(t)|<0.05\) \(\mathrm{cm}\) for all \(t>\tau\)

A cubic block of side \(l\) and mass density \(\rho\) per unit volume is floating in a fluid of mass density \(\rho_{0}\) per unit volume, where \(\rho_{0}>\rho .\) If the block is slightly depressed and then released, it oscillates in the vertical direction. Assuming that the viscous damping of the fluid and air can be neglected, derive the differential equation of motion and determine the period of the motion. Hint Use archimedes' principle: An object that is completely or partially submerged in a fluid is acted on by an upward (bouyant) equal to the weight of the displaced fluid.

Use the method outlined in Problem 28 to solve the given differential equation. $$ t^{2} y^{\prime \prime}-2 t y^{\prime}+2 y=4 t^{2}, \quad t>0 ; \quad y_{1}(t)=t $$
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