Chapter 4: Problem 3
Use the method of variation of parameters to determine the general solution of the given differential equation. $$ y^{\prime \prime \prime}-2 y^{\prime \prime}-y^{\prime}+2 y=e^{4 t} $$
Short Answer
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Question: Determine the general solution for the given third-order linear differential equation using the method of variation of parameters: $$y^{\prime \prime \prime}-2 y^{\prime \prime}-y^{\prime}+2 y=e^{4t}$$
Answer: The general solution for the given third-order linear differential equation using the method of variation of parameters is: $$y(t) = C_1e^{-t} + C_2e^{t} + C_3e^{2t} + (-5t)e^{-t} - te^{t}$$, where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants.
Step by step solution
01
Characteristic Equation of the Homogeneous Equation
First, let's find the homogeneous equation related to the given differential equation:
$$
y^{\prime \prime \prime}-2 y^{\prime \prime}-y^{\prime}+2 y=0
$$
To find the characteristic equation, we will replace each derivative with an exponent. Let's assume a solution of the form \(y=e^{rt}\) for the homogeneous equation. For the given equation, the characteristic equation becomes:
$$
r^3 - 2r^2 - r + 2 = 0
$$
02
Finding the Roots of the Characteristic Equation
The characteristic equation is a cubic equation. We should try to find at least one real root which will help factorize the equation into smaller polynomials to find the remaining roots. By trial and error or synthetic division, one will observe that \(r=1\) is a root.
Now we can factor the equation as:
$$
(r-1)(r^2-r-2)=0
$$
The quadratic equation has roots:
$$
r_{1,2}=\frac{1\pm \sqrt{1^2-4(-2)}}{2}=\frac{1\pm \sqrt{9}}{2}=\frac{1\pm 3}{2}
$$
Thus, the roots are \(r_1=-1\), \(r_2=1\), and \(r_3=2\).
03
Finding the Complementary Function
Using the roots \(-1\), \(1\), and \(2\), we can write the complementary function for the given homogeneous equation as:
$$
y_c = C_1e^{-t} + C_2e^{t} + C_3e^{2t}
$$
where \(C_1, C_2\), and \(C_3\) are constants.
04
Applying Variation of Parameters
Next, let's use the method of variation of parameters to find the particular solution. We need to take the first derivative of the complementary solution:
$$
y_c'= -C_1e^{-t} + C_2e^{t} + 2C_3e^{2t}
$$
To find the particular solution, we will assume it in the form:
$$
y_p =u_1e^{-t}+u_2e^{t}+u_3e^{2t}
$$
Now, we need to differentiate \(y_p\):
$$
y_p' = -u_1'e^{-t}+u_1e^{-t}+u_2'e^{t}+u_2e^{t}+ 2u_3'e^{2t} +2u_3e^{2t}
$$
We will differentiate \(y_p'\) once more to get:
$$
y_p'' = u_1'e^{-t}+2u_1e^{-t}+u_2'e^{t}+2u_2e^{t}+ 4u_3'e^{2t} +4u_3e^{2t}
$$
Now we take the third derivative of \(y_p\):
$$
y_p''' = -u_1'e^{-t}+3u_1e^{-t}+u_2'e^{t}+3u_2e^{t}+ 8u_3'e^{2t} +8u_3e^{2t}
$$
05
Determining the Particular Solution
Now let's substitute \(y_p'\)', \(y_p''\), and \(y_p'''\) into the given differential equation:
$$
y_p''' - 2y_p'' - y_p' + 2y_p = e^{4t}
$$
Substitute:
$$
(-u_1'e^{-t}+3u_1e^{-t}+u_2'e^{t}+3u_2e^{t}+ 8u_3'e^{2t} +8u_3e^{2t})-2(u_1'e^{-t}+2u_1e^{-t}+u_2'e^{t}+2u_2e^{t}+ 4u_3'e^{2t} +4u_3e^{2t}) - (-u_1'e^{-t}+u_1e^{-t}+u_2'e^{t}+u_2e^{t}+ 2u_3'e^{2t} +2u_3e^{2t}) +2(u_1e^{-t}+u_2e^{t}+u_3e^{2t}) = e^{4t}
$$
Simplify and collect terms:
$$
(-u_1'e^{-t}+5u_1e^{-t} - u_2'e^{t}+u_2e^{t}+ 6u_3'e^{2t} +10u_3e^{2t}) = e^{4t}
$$
Now we compare the coefficients of the exponential of like powers:
$$
-u_1'=5, -u_2'=1, 6u_3'=0
$$
Integrating each equation with respect to t yields:
$$
u_1(t) = -5t + C_1, u_2(t) = -t + C_2, u_3(t) = C_3
$$
Now, plugging \(u_1(t)\), \(u_2(t)\), and \(u_3(t)\) back into the expression for \(y_p\):
$$
y_p =(-5t + C_1)e^{-t} + (-t + C_2)e^{t} + C_3e^{2t}
$$
Since \(y_p\) is a particular solution and we want the first function linearly independent of the complementary function, we can take \(C_1 = C_2 = C_3 = 0\). This gives:
$$
y_p = (-5t)e^{-t} - te^{t}
$$
06
General Solution
Now, the general solution is the sum of the complementary and particular solutions:
$$
y(t) = y_c + y_p = C_1e^{-t} + C_2e^{t} + C_3e^{2t} + (-5t)e^{-t} - te^{t}
$$
This is the general solution of the given third-order differential equation using the method of variation of parameters.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are mathematical equations that relate a function to its derivatives. They play a crucial role in modeling various physical systems where rates of change are involved, such as in physics, engineering, and biology. A differential equation describes how a quantity changes with respect to another, often time, providing insights into the behavior of dynamic systems.
In mathematical notation, a differential equation typically involves one or more unknown functions and their derivatives. For example, in the exercise, we have a third-order differential equation:
In mathematical notation, a differential equation typically involves one or more unknown functions and their derivatives. For example, in the exercise, we have a third-order differential equation:
- \(y''' - 2y'' - y' + 2y = e^{4t}\)
Complementary Function
The complementary function is a solution to the homogeneous part of a differential equation. The homogeneous equation is derived by setting the non-homogeneous term (like \(e^{4t}\) in our exercise) to zero. Thus, our homogeneous equation becomes:
- \(y''' - 2y'' - y' + 2y = 0\)
- \(r^3 - 2r^2 - r + 2 = 0\)
- \(y_c = C_1e^{-t} + C_2e^{t} + C_3e^{2t}\)
Particular Solution
The particular solution of a differential equation accounts for the non-homogeneous part, specifically addressing the input function (such as \(e^{4t}\)) that causes deviation from the homogeneous solution. In the method of variation of parameters, we assume a form like the complementary solution but replace the constants with functions:
Through integration, we obtain:
- \(y_p = u_1e^{-t} + u_2e^{t} + u_3e^{2t}\)
Through integration, we obtain:
- \(u_1(t) = -5t + C_1\)
- \(u_2(t) = -t + C_2\)
- \(u_3(t) = C_3\)
Characteristic Equation
The characteristic equation is a crucial tool in solving linear homogeneous differential equations. It arises when you seek exponential solutions to a differential equation with constant coefficients, like:
- \(y''' - 2y'' - y' + 2y = 0\)
- \(r^3 - 2r^2 - r + 2 = 0\)
- Real roots give rise to exponential terms (like \(e^{rt}\)).
- Complex roots lead to oscillatory solutions, which we handle using trigonometric functions.