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Magazine Chapter 4: Problem 20
Find the general solution of each of the differential equations in exercise. $$ \frac{d^{4} y}{d x^{4}}-\frac{d^{3} y}{d x^{3}}-3 \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+2 y=0. $$
Short Answer
Expert verified
The general solution of the given differential equation is:
\[
y(x) = C_1e^x + C_2xe^x + C_3e^{2x} -\frac{3}{2}x^2e^x - \frac{1}{2}xe^x - \frac{1}{2}e^x,
\]
where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants.
Step by step solution
01
Solve the homogeneous equation for the complementary solution
The given differential equation is
\[
\frac{d^3y}{dx^3} - 4\frac{d^2y}{dx^2} + 5\frac{dy}{dx} - 2y = 0,
\]
which is a homogeneous linear differential equation with constant coefficients. To solve this, we assume a solution of the form \(y = e^{\lambda x}\) and substitute it into the equation to find the characteristic equation.
\[
\left(\lambda^3 - 4\lambda^2 + 5\lambda - 2\right)e^{\lambda x} = 0.
\]
Now we need to solve the characteristic equation \( \lambda^3 - 4\lambda^2 + 5\lambda - 2 = 0 \) for its roots.
The roots are \(\lambda_1 = 1\), \(\lambda_2 = 1\), and \(\lambda_3 = 2\). Thus, the complementary solution (general solution of the homogeneous equation) is
\[
y_c(x) = C_1e^x + C_2xe^x + C_3e^{2x},
\]
where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants.
02
Find the particular solution using the variation of parameters method
Since the non-homogeneous term of the given differential equation is of the form \((3x^2 - 7)e^x\), we can assume a particular solution \(y_p(x)\) of the form \(y_p(x) = Ax^2e^x + Bxe^x + Ce^x\), where \(A\), \(B\), and \(C\) are coefficients to be determined.
Now, we compute the first, second, and third derivatives of \(y_p(x)\).
\[
y'_p(x) = 2Axe^x + Ae^x + Be^x + Ce^x,
\]
\[
y''_p(x) = 2Ae^x + 4Axe^x + 2Be^x + 2Ce^x,
\]
\[
y'''_p(x) = 6Ae^x + 8Axe^x + 4Be^x + 4Ce^x.
\]
Substitute these derivatives into the given differential equation, and solve for the coefficients \(A\), \(B\), and \(C\):
\[
(6Ae^x + 8Axe^x + 4Be^x + 4Ce^x) - 4(2Ae^x + 4Axe^x + 2Be^x + 2Ce^x) + 5(2Axe^x + Ae^x + Be^x + Ce^x) - 2(Ax^2e^x + Bxe^x + Ce^x) = (3x^2 - 7)e^x.
\]
Equating the coefficients of the power of \(x\) with the non-homogeneous terms, we obtain a system of equations:
\[
-2A = 3, \quad
-5A + 5B - 2B = -7, \quad
6A - 16A + 20A - 10B + 5C = 0.
\]
Solving this system, we have \(A = -\frac{3}{2}\), \(B = -\frac{1}{2}\), and \(C = -\frac{1}{2}\). Thus, the particular solution is
\[
y_p(x) = -\frac{3}{2}x^2e^x - \frac{1}{2}xe^x - \frac{1}{2}e^x.
\]
03
Combine the complementary solution and the particular solution
Finally, we find the general solution of the given differential equation by adding the complementary solution and the particular solution:
\[
y(x) = y_c(x) + y_p(x) = C_1e^x + C_2xe^x + C_3e^{2x} -\frac{3}{2}x^2e^x - \frac{1}{2}xe^x - \frac{1}{2}e^x,
\]
where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Linear Differential Equations
A homogeneous linear differential equation is an equation that can be written in the form: \[ a_n \frac{d^n y}{dx^n} + a_{n-1} \frac{d^{n-1} y}{dx^{n-1}} + \dots + a_1 \frac{dy}{dx} + a_0 y = 0, \] where \( a_0, a_1, \ldots, a_n \) are constants or functions of \( x \), and \( y \) is the function of \( x \) we want to solve for.This form does not include any terms that are not multiplied by \( y \) or its derivatives.
To solve these equations, we often use methods involving characteristic equations, which involve assuming potential solutions of a particular form and substituting back into the original equation to solve for unknown parameters. Solving these equations provides us with a complementary solution: the solution to the homogeneous part of the differential equation.
To solve these equations, we often use methods involving characteristic equations, which involve assuming potential solutions of a particular form and substituting back into the original equation to solve for unknown parameters. Solving these equations provides us with a complementary solution: the solution to the homogeneous part of the differential equation.
Characteristic Equation
The characteristic equation is critically used when solving homogeneous linear differential equations with constant coefficients.We assume a solution of the form \( y = e^{\lambda x} \) and substitute it back into the differential equation.
In the exercise, the differential equation is \[ \frac{d^3y}{dx^3} - 4\frac{d^2y}{dx^2} + 5\frac{dy}{dx} - 2y = 0. \]By substituting \( y = e^{\lambda x} \), we arrive at the characteristic equation:\[ \lambda^3 - 4\lambda^2 + 5\lambda - 2 = 0. \]
Finding the roots of this polynomial equation allows us to determine the form of the complementary solution.In this problem, the roots were \( \lambda_1 = 1 \), \( \lambda_2 = 1 \), and \( \lambda_3 = 2 \), leading to a solution:\[ y_c(x) = C_1e^x + C_2xe^x + C_3e^{2x}, \]where \( C_1, C_2, C_3 \) are constants and \( xe^x \) appears because of the repeated roots.
In the exercise, the differential equation is \[ \frac{d^3y}{dx^3} - 4\frac{d^2y}{dx^2} + 5\frac{dy}{dx} - 2y = 0. \]By substituting \( y = e^{\lambda x} \), we arrive at the characteristic equation:\[ \lambda^3 - 4\lambda^2 + 5\lambda - 2 = 0. \]
Finding the roots of this polynomial equation allows us to determine the form of the complementary solution.In this problem, the roots were \( \lambda_1 = 1 \), \( \lambda_2 = 1 \), and \( \lambda_3 = 2 \), leading to a solution:\[ y_c(x) = C_1e^x + C_2xe^x + C_3e^{2x}, \]where \( C_1, C_2, C_3 \) are constants and \( xe^x \) appears because of the repeated roots.
Particular Solution
The particular solution of a differential equation can sometimes be found using varied methods, such as undetermined coefficients or variation of parameters.For non-homogeneous differential equations, a particular solution accounts for the presence of an external non-zero function on the right-hand side of the equation.
In this specific exercise, the equation given is:\[ \frac{d^3y}{dx^3} - 4\frac{d^2y}{dx^2} + 5\frac{dy}{dx} - 2y = 3x^2 e^x - 7e^x. \]
We assumed a particular solution of the form \( y_p(x) = Ax^2e^x + Bxe^x + Ce^x \).By calculating the derivatives and substituting back into the original differential equation, the aim is to solve for the coefficients \( A, B, \) and \( C \).In the exercise, it was found that \( A = -\frac{3}{2} \), \( B = -\frac{1}{2} \), and \( C = -\frac{1}{2} \), which lead to the particular solution:\[ y_p(x) = -\frac{3}{2}x^2e^x - \frac{1}{2}xe^x - \frac{1}{2}e^x. \]
In this specific exercise, the equation given is:\[ \frac{d^3y}{dx^3} - 4\frac{d^2y}{dx^2} + 5\frac{dy}{dx} - 2y = 3x^2 e^x - 7e^x. \]
We assumed a particular solution of the form \( y_p(x) = Ax^2e^x + Bxe^x + Ce^x \).By calculating the derivatives and substituting back into the original differential equation, the aim is to solve for the coefficients \( A, B, \) and \( C \).In the exercise, it was found that \( A = -\frac{3}{2} \), \( B = -\frac{1}{2} \), and \( C = -\frac{1}{2} \), which lead to the particular solution:\[ y_p(x) = -\frac{3}{2}x^2e^x - \frac{1}{2}xe^x - \frac{1}{2}e^x. \]
Variation of Parameters
Variation of parameters is a powerful technique used to find the particular solution of a non-homogeneous differential equation.Unlike the method of undetermined coefficients, variation of parameters can work even when the function on the right-hand side of the equation is complicated or not of a form easy to guess.
The idea is to take the complementary solution and introduce functions into the solution instead of constants.You then solve these functions to satisfy the non-homogeneous part of the differential equation.
Although not explicitly used in the step-by-step solution provided in the exercise, variation of parameters is a technique good to understand.It involves setting the particular solution as \( y_p(x) = u_1(x) y_{1} + u_2(x) y_2 + \ldots \), where \( y_{1}, y_2, \ldots \) are solutions of the complementary equation. You solve for \( u_1(x), u_2(x) \) by matching terms with the non-homogeneous part.
The idea is to take the complementary solution and introduce functions into the solution instead of constants.You then solve these functions to satisfy the non-homogeneous part of the differential equation.
Although not explicitly used in the step-by-step solution provided in the exercise, variation of parameters is a technique good to understand.It involves setting the particular solution as \( y_p(x) = u_1(x) y_{1} + u_2(x) y_2 + \ldots \), where \( y_{1}, y_2, \ldots \) are solutions of the complementary equation. You solve for \( u_1(x), u_2(x) \) by matching terms with the non-homogeneous part.
