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Chapter 3: Problem 25

Solve the initial value problems given in Problems. $$ 3 y^{(3)}+2 y^{\prime \prime}=0 ; y(0)=-1, y^{\prime}(0)=0, y^{\prime \prime}(0)=1 $$

Short Answer

Expert verified
The solution is \( y(t) = -\frac{13}{4} + \frac{3}{2}t + \frac{9}{4} e^{-\frac{2}{3}t} \).

Step by step solution

01

Identifying the Differential Equation

The given differential equation is a third-order homogeneous linear differential equation: \[ 3y^{(3)} + 2y'' = 0. \]
02

Finding the Characteristic Equation

First, we need to find the characteristic equation of the given differential equation. Assume a solution of the form \( y = e^{mt} \). Then the characteristic equation derived from substituting into the differential equation is: \[ 3m^3 + 2m^2 = 0. \]
03

Solving the Characteristic Equation

Factor the characteristic equation:\[ m^2(3m + 2) = 0. \]This gives solutions for \( m \): - \( m_1 = 0 \) (with multiplicity 2)- \( m_3 = -\frac{2}{3} \).
04

Writing the General Solution

Since we have the roots, we can write the general solution for the differential equation as:\[ y(t) = c_1 + c_2 t + c_3 e^{-\frac{2}{3}t}. \]
05

Applying Initial Conditions: Step 1

Apply the initial condition \( y(0) = -1 \) to find \( c_1 \):\[ c_1 + c_3 = -1 \]
06

Applying Initial Conditions: Step 2

Find the first derivative of \( y(t) \):\[ y'(t) = c_2 - \frac{2}{3}c_3 e^{-\frac{2}{3}t}. \]Apply the condition \( y'(0) = 0 \):\[ c_2 - \frac{2}{3}c_3 = 0 \]
07

Applying Initial Conditions: Step 3

Find the second derivative of \( y(t) \):\[ y''(t) = \frac{4}{9}c_3 e^{-\frac{2}{3}t}. \]Use the condition \( y''(0) = 1 \) to find \( c_3 \): \[ \frac{4}{9}c_3 = 1 \]\[ c_3 = \frac{9}{4}. \]
08

Solving for Constants

Substitute \( c_3 = \frac{9}{4} \) back into the equations to find \( c_1 \) and \( c_2 \): \[ c_1 + \frac{9}{4} = -1 \] \[ c_1 = -\frac{13}{4} \] \[ c_2 - \frac{2}{3} \times \frac{9}{4} = 0 \] \[ c_2 = \frac{3}{2}. \]
09

Final Solution

The final solution that satisfies the differential equation and all initial conditions is:\[ y(t) = -\frac{13}{4} + \frac{3}{2}t + \frac{9}{4} e^{-\frac{2}{3}t}. \]

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Key Concepts

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

Initial Value Problem
An initial value problem is a type of differential equation accompanied by specific conditions called initial conditions. These initial conditions state the value of the function and its derivatives at a specific point, usually at the start. In our problem:
  • The differential equation is given as: \( 3y^{(3)} + 2y'' = 0 \).
  • The initial conditions are \( y(0) = -1 \), \( y'(0) = 0 \), and \( y''(0) = 1 \).
Initial conditions are crucial as they help us determine the specific solution that fits this scenario, instead of a general solution with multiple constants. By applying these conditions, we can identify the values of the constants in the general solution and find the unique solution that meets all the conditions.
Linear Differential Equations
Linear differential equations are equations involving derivatives in which the dependent variable and its derivatives appear linearly. In simple terms:
  • Each term in the equation is either a derivative of the variable or a product of a constant and such derivatives.
  • There are no powers or non-linear functions of derivatives irregularly mixed or composed.
In this exercise, our differential equation is linear because each term involves a derivative of \( y \) multiplied by a constant: \( 3y^{(3)} \) and \( 2y'' \). Linear equations are preferred in many applications because they are simpler to solve and provide a very clear view of the dynamic behavior of systems.
Characteristic Equation
The characteristic equation is derived from a linear differential equation and helps find solutions by turning a continuous problem into an algebraic one. To find it:
  • Assume a solution of the form \( y=e^{mt} \), which when substituted into the differential equation yields the characteristic equation.
  • In our specific case, substituting gives \( 3m^3 + 2m^2 = 0 \).
By solving the characteristic equation, we can find the roots \( m \) that correspond to the solutions of the differential, as exponentials or polynomials. This method transforms solving a differential equation into the simpler task of factoring a polynomial.
Homogeneous Differential Equation
A homogeneous differential equation is a type of differential equation where every term is a derivative of the unknown function, and there's no standalone function or separate constant term. This specific exercise shows:
  • The differential equation \( 3y^{(3)} + 2y'' = 0 \) does not include any terms without \( y \) derivatives.
Solutions to homogeneous equations generally take forms based on exponential solutions of the characteristic equation. For example, with the roots \( m=0 \) and \( m=-\frac{2}{3} \) from the characteristic equation, solutions are made up of terms like constants \( c_1 \), \( c_2 t \), and \( c_3 e^{- rac{2}{3}t} \). By using these solutions with initial conditions, we find specific values for these constants.

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Most popular questions from this chapter

Use the method of variation of parameters to find a particular solution of the given differential equation. $$ y^{\prime \prime}+9 y=2 \sec 3 x $$

Find a particular solution \(y_{p}\) of the given equation. In all these problems, primes denote derivatives with respect to \(x\). $$ y^{(3)}-y=e^{x}+7 $$

Each of Problems 15 through 18 gives the parameters for a forced mass-spring- dashpot system with equation \(m x^{\prime \prime}+c x^{\prime}+\) \(k x=F_{0} \cos \omega t .\) Investigate the possibility of practical resonance of this system. In particular, find the amplitude \(C(\omega)\) of steady periodic forced oscillations with frequency \(\omega .\) Sketch the graph of \(C(\omega)\) and find the practical resonance frequency \(\omega(\) if any). $$ m=1, c=2, k=2, F_{0}=2 $$

Make the substitution \(v=\ln x\) of Problem 51 to find general solutions (for \(x>0\) ) of the Euler equations in Problems. $$ x^{2} y^{\prime \prime}+x y^{\prime}+9 y=0 $$

In each of Problems 11 through 14, find and plot both the steady periodic solution \(x_{\mathrm{sp}}(t)=C \cos (\omega t-\alpha)\) of the given differential equation and the transient solution \(x_{\mathrm{tr}}(t)\) that satisfies the given initial conditions. $$ x^{\prime \prime}+4 x^{\prime}+5 x=10 \cos 3 t ; x(0)=x^{\prime}(0)=0 $$
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