Problem 15 Find the general solution of the... [FREE SOLUTION] (2024)

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Chapter 9: Problem 15

Find the general solution of the system of equations. \(x^{\prime}=y+2 t+3, y^{\prime}=-x+4 t-2\)

Short Answer

Expert verified

General solutions are \( x(t) = C_1 \cos(t) + C_2 \sin(t) + 4t \) and \( y(t) = -C_1 \sin(t) + C_2 \cos(t) - 2t + 1 \).

Step by step solution

Write down the system of differential equations

Consider the system given:\( x' = y + 2t + 3 \)\( y' = -x + 4t - 2 \)

Differentiate the first equation

Differentiate the first equation with respect to \(t\):\( x'' = y' + 2 \)

Substitute for \( y' \) from the second equation

Substitute \( y' = -x + 4t - 2 \) from the second equation into the differentiated equation:\( x'' = -x + 4t - 2 + 2 \)This simplifies to \( x'' = -x + 4t \)

Solve the hom*ogeneous equation

Solve the hom*ogeneous part of the equation \( x'' + x = 0 \). The characteristic equation is:\( r^2 + 1 = 0 \)The roots are \( r = \pm i \), hence the general solution for the hom*ogeneous part is:\( x_h(t) = C_1 \cos(t) + C_2 \sin(t) \)

Solve the non-hom*ogeneous equation

Use the method of undetermined coefficients to solve for the particular solution. Assume a particular solution of the form \( x_p(t) = At + B \). Substitute this into the non-hom*ogeneous equation:\( (At + B)'' + (At + B) = 4t \)This simplifies to:\( A = 4 \), \( B = 0 \)Thus, the particular solution is \( x_p(t) = 4t \)

Formulate the general solution for x

Combine the hom*ogeneous and particular solutions to get the general solution for \( x(t) \):\( x(t) = C_1 \cos(t) + C_2 \sin(t) + 4t \)

Find the solution for y

To find \( y(t) \), substitute \( x(t) \) back into the original equation \( x' = y + 2t + 3 \):Differentiating \( x(t) \) gives:\( x'(t) = -C_1 \sin(t) + C_2 \cos(t) + 4 \)Thus, \( -C_1 \sin(t) + C_2 \cos(t) + 4 = y + 2t + 3 \)Solving for \( y \):\( y(t) = -C_1 \sin(t) + C_2 \cos(t) + 4 - 2t - 3 \)Which simplifies to:\( y(t) = -C_1 \sin(t) + C_2 \cos(t) - 2t + 1 \)

Key Concepts

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

System of Differential Equations

A system of differential equations consists of multiple differential equations that relate various functions and their derivatives. In the given problem, we have:

\( x' = y + 2t + 3 \)
\( y' = -x + 4t - 2 \)

This set involves two dependent variables, \( x \) and \( y \), each changing over time \( t \). Systems can be solved using various methods including substitution and matrix approaches. The key is to find expressions for \( x(t) \) and \( y(t) \) that satisfy both equations.

hom*ogeneous Equations

A hom*ogeneous differential equation is one where all the terms are functions of the dependent variable and its derivatives only.
In our exercise, the equation \( x'' + x = 0 \) is hom*ogeneous. The characteristic equation \( r^2 + 1 = 0 \) gives roots \( r = \pm i \). Using these roots, the general solution for the hom*ogeneous part is:
\( x_h(t) = C_1 \cos(t) + C_2 \sin(t) \)
hom*ogeneous solutions typically involve exponentials, sines, or cosines depending on the nature of the roots of the characteristic equation.

Non-hom*ogeneous Equations

Non-hom*ogeneous equations include extra terms that are not functions of the dependent variable and its derivatives.
For example, in our problem \( x'' = -x + 4t \), the \( 4t \) term makes it non-hom*ogeneous. Solutions to such equations are the sum of the hom*ogeneous solution and a particular solution. To solve, we often guess the form of the particular solution based on the non-hom*ogeneous term structure.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique for finding the particular solution of non-hom*ogeneous differential equations.
We assume a form for the particular solution with unknown coefficients, then determine those coefficients by substituting back into the original equation.
In our problem, we assume \( x_p(t) = At + B \). Substituting this into the non-hom*ogeneous equation \( x'' = -x + 4t \):

\( (At + B)'' + (At + B) = 4t \)
Simplifying gives \( A = 4 \), \( B = 0 \).

So, \( x_p(t) = 4t \).

Particular Solution

A particular solution of a non-hom*ogeneous differential equation specifically satisfies the entire equation.
Having found the hom*ogeneous solution, we add the particular solution to form the complete solution.
In our problem, combining the hom*ogeneous solution \( x_h(t) = C_1 \cos(t) + C_2 \sin(t) \) with the particular solution \( x_p(t) = 4t \) gives the general solution: \( x(t) = C_1 \cos(t) + C_2 \sin(t) + 4t \)
This comprehensive solution accounts for all possible behaviors of the system under the given differential equations.

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Most popular questions from this chapter

Solve. \(y^{\prime \prime}-4 y^{\prime}+5 y=5 x+1\)Let \(x\) and \(y\) represent the populations (in thousands) of prey and predatorsthat share a habitat. For the given system of differential equations, find andclassify the equilibrium points. \(x^{\prime}(t)=0.5 x-0.2 x y, y^{\prime}(t)=-0.4 y+0.1 x y\)Find the general solution of the system of equations. \(x^{\prime}=2 y, y^{\prime}=-18 x\)Find the equilibrium points and assess their stability. \(x^{\prime}(t)=0.8 x(1-x / 24)-0.2 x y\) \(y^{\prime}(t)=-0.6 y+0.1 x y\)Rewrite the system of differential equations into matrix form. \(x^{\prime}=-x+3 y, y^{\prime}=3 x+y\)

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