Chapter 3: Problem 1
Solve the given system of differential equations by systematic elimination. $$ \begin{aligned} &\frac{d x}{d t}=2 x-y \\ &\frac{d y}{d t}=x \end{aligned} $$
Short Answer
Expert verified
The general solutions are \( y(t) = (C_1 + C_2 t) e^t \) and \( x(t) = (C_1 + C_2 + C_2 t) e^t \).
Step by step solution
01
Find the Second Derivative of y
Start by differentiating the second equation with respect to \( t \): \[ \frac{d}{dt} \left( \frac{d y}{dt} \right) = \frac{d}{dt}(x) \] This equation becomes \( \frac{d^2 y}{dt^2} = \frac{dx}{dt} \). From the first equation, \( \frac{dx}{dt} = 2x - y \). Thus, we have \( \frac{d^2 y}{dt^2} = 2x - y \).
02
Express x in Terms of y and its Derivatives
From the second original equation, \( \frac{d y}{dt} = x \), isolate \( x \): \[ x = \frac{d y}{dt} \]
03
Substitute x into the Second Derivative Equation
Substitute \( x = \frac{d y}{dt} \) into the equation for \( \frac{d^2 y}{dt^2} \): \[ \frac{d^2 y}{dt^2} = 2 \left( \frac{d y}{dt} \right) - y \] This gives us the second-order differential equation: \[ \frac{d^2 y}{dt^2} - 2 \frac{d y}{dt} + y = 0 \]
04
Solve the Second-Order Differential Equation
This is a homogeneous linear differential equation with constant coefficients. Assume a solution of the form \( y = e^{\lambda t} \). Substitute into the differential equation: \[ \lambda^2 e^{\lambda t} - 2 \lambda e^{\lambda t} + e^{\lambda t} = 0 \] Factor out \( e^{\lambda t} \):\[ e^{\lambda t} ( \lambda^2 - 2\lambda + 1 ) = 0 \] Solve for \( \lambda \) by setting the quadratic to zero: \[ \lambda^2 - 2\lambda + 1 = (\lambda - 1)^2 = 0 \] Thus, \( \lambda = 1 \) is a double root.
05
Write the General Solution for y
With a double root from the characteristic equation, the general solution for \( y \) is: \[ y(t) = (C_1 + C_2 t) e^{t} \] where \( C_1 \) and \( C_2 \) are arbitrary constants.
06
Determine the General Solution for x
Using \( x = \frac{d y}{dt} \), differentiate \( y(t) \) to find \( x \): \[ \frac{d y}{dt} = C_1 e^{t} + C_2 e^{t} + C_2 t e^{t} = (C_1 + C_2 + C_2 t) e^{t} \] Hence, \( x(t) = (C_1 + C_2 + C_2 t) e^{t} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Linear Differential Equation
A homogeneous linear differential equation is a type of differential equation in which every term is a function of the dependent variable and its derivatives, and all terms add up to zero. These equations often take the form:
\[ a_n \frac{d^n y}{dt^n} + a_{n-1} \frac{d^{n-1} y}{dt^{n-1}} + \.\.\. + a_1 \frac{dy}{dt} + a_0 y = 0 \]
where the coefficients \(a_n, a_{n-1}, ..., a_0\) are constants. The term `homogeneous` implies that if the functions on the left-hand side are zero, the right-hand side of the equation is zero.
\[ a_n \frac{d^n y}{dt^n} + a_{n-1} \frac{d^{n-1} y}{dt^{n-1}} + \.\.\. + a_1 \frac{dy}{dt} + a_0 y = 0 \]
where the coefficients \(a_n, a_{n-1}, ..., a_0\) are constants. The term `homogeneous` implies that if the functions on the left-hand side are zero, the right-hand side of the equation is zero.
- Each term in a homogeneous equation can be derived from solving the characteristic equation.
- These types of equations can often be solved using methods related to linear algebra or polynomial factorization.
Characteristic Equation
The characteristic equation is a fundamental tool in solving linear differential equations with constant coefficients. It is derived from the differential equation by substituting an exponential function as a potential solution.
For a differential equation:\[ a_n \frac{d^n y}{dt^n} + a_{n-1} \frac{d^{n-1} y}{dt^{n-1}} + \.\.\. + a_1 \frac{dy}{dt} + a_0 y = 0 \]
the characteristic equation is:\[ a_n \lambda^n + a_{n-1} \lambda^{n-1} + \.\.\. + a_1 \lambda + a_0 = 0 \]
For a differential equation:\[ a_n \frac{d^n y}{dt^n} + a_{n-1} \frac{d^{n-1} y}{dt^{n-1}} + \.\.\. + a_1 \frac{dy}{dt} + a_0 y = 0 \]
the characteristic equation is:\[ a_n \lambda^n + a_{n-1} \lambda^{n-1} + \.\.\. + a_1 \lambda + a_0 = 0 \]
- Solving this polynomial gives the roots \(\lambda\) that determine the behavior of the solutions.
- If the roots are distinct, the solution forms are simple exponentials.
- In our case, the double root \(\lambda = 1\) leads to a solution involving extra polynomial terms.
Arbitrary Constants
Arbitrary constants are constants whose values can be chosen freely to fit particular boundary conditions or initial conditions of a differential equation. They arise naturally when integrating differential equations.
In a system of differential equations, after solving for the general form, the solution typically includes several arbitrary constants,
In a system of differential equations, after solving for the general form, the solution typically includes several arbitrary constants,
- These constants can be seen as placeholders for specific conditions like initial velocity or starting amounts in chemical reactions.
- For example, in the solution for \(y(t) = (C_1 + C_2 t) e^{t}\), \(C_1\) and \(C_2\) are constants adjusted to match initial or boundary conditions.
- Their flexibility makes the solution wide-ranging and adaptable to any number of practical problems.
Constant Coefficients
Constant coefficients refer to the coefficients of the derivatives in a differential equation that remain unchanged. In a linear differential equation, these coefficients are linked to the system's physical properties which do not change over time.
For instance, in our differential equation:\[ \frac{d^2 y}{dt^2} - 2 \frac{d y}{dt} + y = 0 \]
the coefficients are the numbers in front of the derivatives like \(-2\), which remain constant.
For instance, in our differential equation:\[ \frac{d^2 y}{dt^2} - 2 \frac{d y}{dt} + y = 0 \]
the coefficients are the numbers in front of the derivatives like \(-2\), which remain constant.
- Equations with constant coefficients are often easier to solve using algebraic methods, such as the characteristic equation.
- They provide consistent behavior over time, which is useful in predicting long-term trends.
- These solutions usually involve exponential functions, as these functions naturally arise from constant coefficient equations.