Chapter 10: Problem 7
Find the general solution to the differential equation. $$y^{\prime \prime}+y^{\prime}-6 y=e^{-3 t}$$
Short Answer
Expert verified
The general solution is \(y(t) = C_1 e^{2t} + C_2 e^{-3t} - \frac{1}{5} t e^{-3t}\).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is \(y'' + y' - 6y = e^{-3t}\). This is a second-order linear non-homogeneous differential equation with constant coefficients.
02
Solve the Homogeneous Equation
First, solve the associated homogeneous equation \(y'' + y' - 6y = 0\). Assume a solution of the form \(y_h = e^{rt}\) and substitute it into the homogeneous part, resulting in the characteristic equation: \(r^2 + r - 6 = 0\).
03
Find the Roots of the Characteristic Equation
Solve \(r^2 + r - 6 = 0\) using the quadratic formula, \(r = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\), where \(a = 1\), \(b = 1\), \(c = -6\). Calculate \(r = \frac{{-1 \pm \sqrt{{1 + 24}}}}{2}\), giving roots \(r = 2\) and \(r = -3\).
04
Write the General Solution for the Homogeneous Equation
Since the roots of the characteristic equation are \(r_1 = 2\) and \(r_2 = -3\), the general solution to the homogeneous equation is \(y_h = C_1 e^{2t} + C_2 e^{-3t}\).
05
Solve the Non-Homogeneous Equation
Use the method of undetermined coefficients to solve \(y'' + y' - 6y = e^{-3t}\). Notice \(e^{-3t}\) is a solution of the homogeneous equation, so choose a particular solution of the form \(y_p = At e^{-3t}\).
06
Differentiate the Particular Solution
Differentiate \(y_p = At e^{-3t}\) to find \(y_p'\) and \(y_p''\): - \(y_p' = A e^{-3t} - 3At e^{-3t}\) - \(y_p'' = -6A e^{-3t} + 9At e^{-3t}\).
07
Substitute into the Differential Equation
Substitute \(y_p\), \(y_p'\), and \(y_p''\) into the original equation:\(-6A e^{-3t} + 9At e^{-3t} + A e^{-3t} - 3At e^{-3t} - 6At e^{-3t} = e^{-3t}\).Combine terms to get:\(-5A e^{-3t} = e^{-3t}\).
08
Solve for the Coefficient
Equate coefficients to find \(A\):\(-5A = 1\), giving \(A = -\frac{1}{5}\). Thus, the particular solution is \(y_p = -\frac{1}{5} t e^{-3t}\).
09
Write the General Solution for the Non-Homogeneous Equation
The general solution to the original non-homogeneous differential equation is the sum of the homogeneous and particular solutions: \(y(t) = C_1 e^{2t} + C_2 e^{-3t} - \frac{1}{5} t e^{-3t}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second-order linear differential equations
A second-order linear differential equation is an equation that involves the second derivative of a function. These equations can describe various physical phenomena, like oscillations and wave motion. The general form of these equations is given by:\[ a y'' + b y' + c y = g(t) \]where:
- \( y'' \) is the second derivative of \( y \).
- \( y' \) is the first derivative of \( y \).
- \( a, b, \) and \( c \) are constants.
- \( g(t) \) is a function of \( t \) and represents a non-homogeneous part.
Method of undetermined coefficients
The method of undetermined coefficients is a technique for finding a particular solution to non-homogeneous linear differential equations with constant coefficients. The strategy involves assuming a form for the particular solution, based on the structure of the non-homogeneous term, and then determining the coefficients that make this assumed solution satisfy the equation.For instance, if the non-homogeneous term is of the form \( e^{kt} \), we might suppose a particular solution such as \( y_p = Ae^{kt} \). If the form resembles terms in the complementary solution, we multiply by \( t \) or higher powers of \( t \) until independence is achieved.The steps in this method are:
- Identify the form of the particular solution based on \( g(t) \).
- Substitute this assumed solution into the original differential equation.
- Solve for the unknown coefficients.
Characteristic equation
The characteristic equation is a vital step in solving linear differential equations. Derived by substituting a trial solution of the form \( y_h = e^{rt} \) into the homogeneous part of the differential equation, it provides a polynomial whose roots determine the form of the homogeneous solution. For a differential equation like:\[ y'' + y' - 6y = 0 \]The characteristic equation would be:\[ r^2 + r - 6 = 0 \]We solve this quadratic equation to find the values of \( r \). The nature of the roots (real and distinct, repeated, or complex) will inform the form of the homogeneous solution:
- Real and distinct roots imply solutions of the form \( C_1 e^{r_1 t} + C_2 e^{r_2 t} \).
- Repeated roots result in solutions like \( (C_1 + C_2 t) e^{r t} \).
- Complex roots lead to solutions resembling \( e^{ ext{Re}(r)t} \left( C_1 \, \cos( ext{Im}(r)t) + C_2 \, \sin( ext{Im}(r)t) \right) \).
Particular solution
The particular solution of a differential equation is a specific solution that accounts for the non-homogeneous part of the equation. In the context of the method of undetermined coefficients, once the form is chosen based on the non-homogeneous portion \( g(t) \), the assumed particular solution is substituted back into the original equation.For the equation:\[ y'' + y' - 6y = e^{-3t} \]A suitable particular solution might assume the form \( y_p = At e^{-3t} \) because \( e^{-3t} \) is similar to a homogeneous solution form. The differentiation of this assumed function, followed by substitution into the differential equation, allows us to solve for \( A \). This step ultimately reflects the effect of the term \( g(t) \) within the equation, ensuring the general solution addresses the entire differential equation.
Homogeneous equation
A homogeneous differential equation is one where the non-homogeneous term \( g(t) \) is zero, simplifying the original equation to:\[ a y'' + b y' + c y = 0 \]The focus here is to find the complementary solution, which only accounts for the characteristic behavior of the differential equation without external forces or inputs.Using our earlier example:\[ y'' + y' - 6y = 0 \]Assume \( y_h = e^{rt} \), leading to the characteristic equation \( r^2 + r - 6 = 0 \). Solving this provides the roots, indicating how the system naturally behaves. The solutions to this equation generally involve exponential terms based on these roots, combining them with arbitrary constants \( C_1 \) and \( C_2 \), which get determined later by initial conditions. This homogeneous solution forms the backbone of the general solution of the complete differential equation.