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Convert the following higher-order equations to systems of first-order equations: (a) \(Y^{\prime \prime \prime}+4 Y^{\prime \prime}+5 Y^{\prime}+2 Y=2 x^{2}+10 x+8\), \(Y(0)=1, \quad Y^{\prime}(0)=-1, \quad Y^{\prime \prime}(0)=3\) The true solution is \(Y(x)=e^{-x}+x^{2}\). (b) \(Y^{\prime \prime}+4 Y^{\prime}+13 Y=40 \cos (x), \quad Y(0)=3, \quad Y^{\prime}(0)=4\) The true solution is \(Y(x)=3 \cos (x)+\sin (x)+e^{-2 x} \sin (3 x)\).

Short Answer

Expert verified
(a) \[\begin{cases} y_1' = y_2 \ y_2' = y_3 \ y_3' = 2x^2 + 10x + 8 - 4y_3 - 5y_2 - 2y_1 \ \text{with } y_1(0) = 1, y_2(0) = -1, y_3(0) = 3 \ \end{cases}\](b) \[\begin{cases} y_1' = y_2 \ y_2' = 40\cos(x) - 4y_2 - 13y_1 \ \text{with } y_1(0) = 3, y_2(0) = 4 \ \end{cases}\]}

Step by step solution

01

- Define the new variables for part (a)

Let us define the new variables to convert the third-order differential equation to a system of first-order equations: \[y_1 = Y, \quad y_2 = Y', \quad y_3 = Y''\]
02

- Write the equations for part (a)

From the definitions, the following first-order differential equations are derived:\[y_1' = y_2\]\[y_2' = y_3\]The original differential equation can be written in terms of these new variables:\[y_3' + 4y_3 + 5y_2 + 2y_1 = 2x^2 + 10x + 8\]
03

- Write the system for part (a)

Now compile the first-order system:\[y_1' = y_2\]\[y_2' = y_3\]\[y_3' = 2x^2 + 10x + 8 - 4y_3 - 5y_2 - 2y_1\]Initial conditions:\[y_1(0) = 1, \quad y_2(0) = -1, \quad y_3(0) = 3\]
04

- Define new variables for part (b)

Similarly, define the new variables to convert the second-order differential equation to a first-order system:\[y_1 = Y, \quad y_2 = Y'\]
05

- Write the equations for part (b)

From the definitions, the following first-order differential equations are derived:\[y_1' = y_2\]The original differential equation can be written in terms of these new variables:\[y_2' + 4y_2 + 13y_1 = 40\cos(x)\]
06

- Write the system for part (b)

Compile the first-order system:\[y_1' = y_2\]\[y_2' = 40\cos(x) - 4y_2 - 13y_1\]Initial conditions:\[y_1(0) = 3, \quad y_2(0) = 4\]

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Key Concepts

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

first-order systems
Converting higher-order differential equations into first-order systems is a common technique in solving complex problems. This method involves breaking down a higher-order equation into multiple first-order equations.
An equation's order is determined by the highest derivative present. For example, a third-order equation involves up to third derivatives. By introducing new variables to represent these derivatives, you can simplify the problem.
Consider the given problem: we have higher-order differential equations of orders three and two, respectively. To transform them into first-order systems, follow these steps:
  • Introduce new variables for each derivative. For example, for the equation with a third derivative, define:
    \text{\(y_1 = Y\)}, \(y_2 = Y'\), and \(y_3 = Y''\).
  • Rewrite the higher-order equation using these new variables.
  • Establish first-order differential equations from the definitions and the original equation.
  • After this process, you will have transformed the problem into a more manageable system of first-order differential equations.
initial conditions
Initial conditions are critical when solving differential equations. They provide the specific values of the function and its derivatives at a given point, usually when \(x = 0\).
In this exercise, initial conditions for part (a) are \(Y(0) = 1\), \(Y'(0) = -1\), and \(Y''(0) = 3\). These conditions ensure a unique solution. For part (b), the initial conditions are \(Y(0) = 3\) and \(Y'(0) = 4\).
These initial conditions translate directly to the values of the newly defined variables at \(x = 0\). For example:
  • For part (a): \(y_1(0) = 1\), \(y_2(0) = -1\), and \(y_3(0) = 3\).
  • For part (b): \(y_1(0) = 3\) and \(y_2(0) = 4\).
By incorporating initial conditions, you can solve the first-order systems accurately. They help in determining the constants of integration and ensuring the solution matches the problem's specifics.
step-by-step solutions
Step-by-step solutions are essential for understanding complex mathematical problems. Breaking down the problem into manageable steps makes it easier to follow and solve efficiently.
Consider the solution process for the given exercise:
  • Step 1: Define new variables for each derivative to simplify the higher-order equation into a first-order system. For part (a), we define: \(y_1 = Y\), \(y_2 = Y'\), and \(y_3 = Y''\).
  • Step 2: Use these variables to rewrite the original higher-order differential equation in terms of the new variables. For example, \(Y^{\text{'''}} + 4Y'' + 5Y' + 2Y = 2x^2 + 10x + 8\) becomes \(y_3' + 4y_3 + 5y_2 + 2y_1 = 2x^2 + 10x + 8\).
  • Step 3: Write the resulting system of first-order equations. For instance, \(y_1' = y_2\), \(y_2' = y_3\), and \(y_3' = 2x^2 + 10x + 8 - 4y_3 - 5y_2 - 2y_1\).
  • Step 4: Apply the initial conditions to these new variables.
Following these steps ensures a clear and structured approach to solving differential equations.
differential equations
Differential equations involve functions and their derivatives. They are fundamental in modeling numerous phenomena in science and engineering.
In the given exercise, we deal with higher-order differential equations. For example, \(Y^{\text{'''}} + 4Y'' + 5Y' + 2Y = 2x^2 + 10x + 8\) and \(Y'' + 4Y' + 13Y = 40\text{cos}(x)\). These equations describe how a function and its derivatives relate to each other.
To solve these equations, especially when dealing with higher-order ones, converting them into a system of first-order differential equations is beneficial. It simplifies the process and makes numerical solutions more feasible.
Understanding the basic principles of differential equations is crucial. They help in predicting the behavior of various systems, from mechanical vibrations to electrical circuits. By learning to convert and solve these equations, one can unlock a deeper understanding of dynamic systems and their behaviors.

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Most popular questions from this chapter

Consider the differential equation $$ Y^{\prime}(x)=f_{1}(x) f_{2}(Y(x)) $$ for some given functions \(f_{1}(x)\) and \(f_{2}(z) .\) This is called a separable differential equation, and it can be solved by direct integration. Write the equation as $$ \frac{Y^{\prime}(x)}{f_{2}(Y(x))}=f_{1}(x) $$ and find the antiderivative of each side: $$ \int \frac{Y^{\prime}(x) d x}{f_{2}(Y(x))}=\int f_{1}(x) d x $$ On the left side, change the integration variable by letting \(z=Y(x)\). Then the equation becomes $$ \int \frac{d z}{f_{2}(z)}=\int f_{1}(x) d x $$ After integrating, replace \(z\) by \(Y(x)\); then solve for \(Y(x)\), if possible. If these integrals can be evaluated, then the differential equation can be solved. Do so for the following problems, finding the general solution and the solution satisfying the given initial condition: \(\begin{array}{ll}\text { (a) } Y^{\prime}(x)=x / Y(x), & Y(0)=2\end{array}\) (b) \(Y^{\prime}(x)=x e^{-Y(x)}, \quad Y(1)=0\) (c) \(\quad Y^{\prime}(x)=Y(x)[a-Y(x)], \quad Y(0)=a / 2, \quad a>0\)

Verify that any function of the form \(Y(x)=c_{1} e^{x}+c_{2} e^{-x}\) satisfies the equation $$ Y^{\prime \prime}(x)-Y(x)=0 $$ Determine \(c_{1}\) and \(c_{2}\) for the function \(Y(x)\) to satisfy the following boundary conditions: (a) \(Y(0)=1, \quad Y(1)=0\); (b) \(Y(0)=1, \quad Y^{\prime}(1)=0\); (c) \(Y^{\prime}(0)=1, \quad Y(1)=0\)

Consider the motion of a particle of mass in falling vertically under the earth's gravitational field, and suppose the downward motion is opposed by a frictional force \(p(v)\) dependent on the velocity \(v(t)\) of the particle. Then the velocity satisfies the equation $$ m v^{\prime}(t)=-m g+p(v), \quad t \geq 0, \quad v(0) \text { given } $$ Let \(m=1 \mathrm{~kg}, g=9.8 \mathrm{~m} / \mathrm{sec}^{2}\), and \(v(0)=0\). Solve the differential equation for \(0 \leq t \leq 20\) and for the following choices of \(p(v)\) : (a) \(p(v)=-0.1 v\), which is positive for a falling body. (b) \(\quad p(v)=0.1 v^{2}\). Find answers to at least three digits of accuracy. Graph the functions \(v(t) .\) Compare the solutions.

Verify that any function of the form \(Y(x)=c_{1} x+c_{2} x^{2}\) satisfies the equation $$ x^{2} Y^{\prime \prime}(x)-2 x Y^{\prime}(x)+2 Y(x)=0 $$ Determine the solution of the equation with the boundary conditions $$ Y(1)=0, \quad Y(2)=1 $$ Use the MATLAB program ODEBVP to solve the boundary value problem for \(h=\) \(0.1,0.05,0.025\), print the errors of the numerical solutions at \(x=1.2,1.4,1.6\) and 1.8. Comment on how errors decrease when \(h\) is halved. Do the same for the extrapolated solutions.

Convert the following system of second-order equations to a larger system of firstorder equations. It arises from studying the gravitational attraction of one mass by another. $$ x^{\prime \prime}(t)=\frac{-c x(t)}{r(t)^{3}}, \quad y^{\prime \prime}(t)=\frac{-c y(t)}{r(t)^{3}}, \quad z^{\prime \prime}(t)=\frac{-c z(t)}{r(t)^{3}} $$ with \(c\) positive constant and \(r(t)=\left[x(t)^{2}+y(t)^{2}+z(t)^{2}\right]^{1 / 2}, t=\) time.

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