Chapter 3: Problem 11
For the given differential equation, $$ 2 y^{\prime \prime}-5 y^{\prime}+2 y=-6 e^{t / 2} $$
Short Answer
Expert verified
Answer: The general solution of the given differential equation is \(y(t) = C_1 e^{t/2} + C_2 e^{2t} + 2te^{t/2}\), where \(C_1\) and \(C_2\) are arbitrary constants.
Step by step solution
01
Find the complementary function (homogeneous equation)
To find the complementary function, we need to write down the homogeneous equation associated with the given differential equation:
$$
2y'' - 5y' + 2y = 0
$$
Next, we will solve this homogeneous equation which is a second-order linear differential equation with constant coefficients. To do this, we will consider the characteristic equation:
$$
2r^2 - 5r + 2 = 0
$$
This is a quadratic equation that can be factored as:
$$
(2r - 1)(r - 2) = 0
$$
Solving for r, we get two roots: r1 = 1/2 and r2 = 2. Therefore, the complementary function is:
$$
y_c(t) = C_1 e^{t/2} + C_2 e^{2t}
$$
02
Find the particular solution
To find the particular solution (y_p) of the given non-homogeneous differential equation, we will use the method of undetermined coefficients.
We are given the non-homogeneous term as \(-6 e^{t/2}\). Observing the right side, we can guess a particular solution of the form:
$$
y_p(t) = Ate^{t/2}
$$
We assume this form because \(e^{t/2}\) is already part of the complementary function, and by multiplying it with \(t\), we ensure linear independence.
Now let's find the first and second derivatives of \(y_p(t)\):
$$
y_p'(t) = A \left( e^{t/2} + \dfrac{t}{2} e^{t/2} \right)
$$
and
$$
y_p''(t) = A \left( \dfrac{1}{4} e^{t/2} + e^{t/2} + \dfrac{t}{2} e^{t/2} \right)
$$
Now substitute \(y_p(t)\), \(y_p'(t)\), and \(y_p''(t)\) into the given differential equation:
$$
2 \left( A \left( \dfrac{1}{4} e^{t/2} + e^{t/2} + \dfrac{t}{2} e^{t/2} \right) \right) - 5 \left( A \left( e^{t/2} + \dfrac{t}{2} e^{t/2} \right) \right) + 2 \left( A te^{t/2} \right) = -6 e^{t/2}
$$
Simplify and collect terms:
$$
A \left( \dfrac{1}{2} e^{t/2} - 3 e^{t/2} + 2 te^{t/2} \right) = -6 e^{t/2}
$$
Now, equating the coefficients of like terms, we get:
$$
A \left( \dfrac{1}{2} - 3 \right) = -6
$$
Solving for A:
$$
A = 2
$$
So, the particular solution is:
$$
y_p(t) = 2te^{t/2}
$$
03
Combine the complementary function and particular solution
The final step is to combine the complementary function, \(y_c(t)\), and the particular solution, \(y_p(t)\), to form the general solution of the given differential equation.
The general solution is given by:
$$
y(t) = y_c(t) + y_p(t) = C_1 e^{t/2} + C_2 e^{2t} + 2te^{t/2}
$$
This is the general solution of the given differential equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complementary Function
The complementary function, often denoted as \(y_c(t)\), is a crucial concept in solving linear homogeneous differential equations. It represents the general solution to the associated homogeneous differential equation, which is obtained by setting the non-homogeneous part to zero. In essence, the complementary function captures the behavior of the system described by the differential equation without the influence of any external forces or inputs.
For the equation \(2 y^{\prime \prime} - 5 y^{\prime} + 2 y = 0\), we derive the complementary function by first determining the characteristic equation \(2r^2 - 5r + 2 = 0\) and finding its roots. Once the roots \(r_1\) and \(r_2\) are determined, the complementary function takes the form \(y_c(t) = C_1 e^{r_1t} + C_2 e^{r_2t}\), where \(C_1\) and \(C_2\) are constants determined by the initial conditions of the problem. This function is a linear combination of solutions that form a basis for the solution space of the homogeneous differential equation.
For the equation \(2 y^{\prime \prime} - 5 y^{\prime} + 2 y = 0\), we derive the complementary function by first determining the characteristic equation \(2r^2 - 5r + 2 = 0\) and finding its roots. Once the roots \(r_1\) and \(r_2\) are determined, the complementary function takes the form \(y_c(t) = C_1 e^{r_1t} + C_2 e^{r_2t}\), where \(C_1\) and \(C_2\) are constants determined by the initial conditions of the problem. This function is a linear combination of solutions that form a basis for the solution space of the homogeneous differential equation.
Particular Solution
Moving beyond the homogeneous case, we are often interested in a specific solution to the non-homogeneous differential equation, fittingly called the particular solution or \(y_p(t)\). When a differential equation includes a distinct external force or input, the particular solution accounts for the response due to that input.
In our exercise, the non-homogeneous term is \( -6 e^{t / 2} \), so we aim to find a function, \(y_p(t)\), that specifically addresses this term. The purpose of \(y_p(t)\) is to plug it into the original non-homogeneous differential equation and achieve balance. The predefined structure of \(y_p(t)\) used in the method of undetermined coefficients should be distinct from the solutions in the complementary function to maintain unique contributions from each part of the general solution.
In our exercise, the non-homogeneous term is \( -6 e^{t / 2} \), so we aim to find a function, \(y_p(t)\), that specifically addresses this term. The purpose of \(y_p(t)\) is to plug it into the original non-homogeneous differential equation and achieve balance. The predefined structure of \(y_p(t)\) used in the method of undetermined coefficients should be distinct from the solutions in the complementary function to maintain unique contributions from each part of the general solution.
Characteristic Equation
To solve second-order linear differential equations with constant coefficients, we leverage the characteristic equation, which is a polynomial whose roots help us construct the complementary function. The equation is derived from the coefficients of the original differential equation and laid out as a characteristic polynomial.
In the example \(2 y^{\prime \prime} - 5 y^{\prime} + 2 y = -6 e^{t / 2}\), the associated characteristic equation is \(2r^2 - 5r + 2 = 0\). The solutions to this quadratic provide us with values of \(r\) that dictate the form of the exponential functions in the complementary function. These solutions are critical as they determine the behavior of the system associated with the homogeneous part of the differential equation.
In the example \(2 y^{\prime \prime} - 5 y^{\prime} + 2 y = -6 e^{t / 2}\), the associated characteristic equation is \(2r^2 - 5r + 2 = 0\). The solutions to this quadratic provide us with values of \(r\) that dictate the form of the exponential functions in the complementary function. These solutions are critical as they determine the behavior of the system associated with the homogeneous part of the differential equation.
Method of Undetermined Coefficients
A practical technique for finding the particular solution of non-homogeneous linear differential equations is the method of undetermined coefficients. This method involves making an educated guess about the form of the particular solution based on the type of the non-homogeneous term.
In our step-by-step solution, the guess was \(y_p(t) = Ate^{t/2}\) since \(e^{t/2}\) is present in the complementary function. By differentiating this form and substituting into the differential equation, we solve for the coefficients that were previously undetermined—in this case, the value of \(A\). The method of undetermined coefficients is straightforward but works best when the non-homogeneous term is a relatively simple function, like a polynomial, exponential, sine, or cosine.
In our step-by-step solution, the guess was \(y_p(t) = Ate^{t/2}\) since \(e^{t/2}\) is present in the complementary function. By differentiating this form and substituting into the differential equation, we solve for the coefficients that were previously undetermined—in this case, the value of \(A\). The method of undetermined coefficients is straightforward but works best when the non-homogeneous term is a relatively simple function, like a polynomial, exponential, sine, or cosine.
Linear Independence
Linear independence is an essential principle that ensures each term in a solution set contributes uniquely to the general solution of a differential equation. For a set of functions to be linearly independent, no function in the set can be written as a linear combination of the others.
In the context of the exercise, ensuring linear independence between the complementary function and the particular solution was vital. While \(e^{t/2}\) was part of the complementary function, we multiplied it by \(t\) in the guess for the particular solution to preserve linear independence. This distinction is crucial for constructing a complete solution that respects the integrity of both the homogeneous and non-homogeneous parts of the equation.
In the context of the exercise, ensuring linear independence between the complementary function and the particular solution was vital. While \(e^{t/2}\) was part of the complementary function, we multiplied it by \(t\) in the guess for the particular solution to preserve linear independence. This distinction is crucial for constructing a complete solution that respects the integrity of both the homogeneous and non-homogeneous parts of the equation.
