Question

For the given differential equation, $$ 2 y^{\prime \prime}-5 y^{\prime}+2 y=t e^{t} $$

Short answer

Question: Solve the differential equation \(2y'' - 5y' + 2y = te^t\). Answer: The general solution of the given differential equation is \(y(x) = C_1 e^{2x} + C_2 e^{\frac{1}{2}x} + te^t + \frac{1}{2}e^t\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Identify the Homogeneous Equation

The given differential equation is: $$ 2y'' - 5y' + 2y = te^t $$ The associated homogeneous equation can be obtained by setting the right side of the equation to zero: $$ 2y'' - 5y' + 2y = 0 $$

Find the Complementary Function

To solve the homogeneous equation, we first need to find the characteristic equation: $$ 2r^2 - 5r + 2 = 0 $$ By factoring, we can obtain: $$ (r-2)(2r-1) = 0 $$ We have two roots, \(r_1 = 2\) and \(r_2 = \frac{1}{2}\). The complementary function (CF) can be expressed as: $$ y_c(x) = C_1 e^{2x} + C_2 e^{\frac{1}{2}x} $$ where \(C_1\) and \(C_2\) are arbitrary constants.

Identify the Particular Integral

To find the particular integral (PI), we try the method of undetermined coefficients. Since the given differential equation is in the form \(te^t\), we will try: $$ y_p(x) = Ate^t + Be^t $$ Now, we need to find the first derivative \(y_p'\) and the second derivative \(y_p''\) of this trial solution: $$ y_p'(x) = Ate^t + Ae^t + Be^t $$ and $$ y_p''(x) = Ate^t + 2Ae^t + Be^t $$ Substitute \(y_p\), \(y_p'\), and \(y_p''\) into the original differential equation: $$ 2(Ate^t + 2Ae^t + Be^t) - 5(Ate^t + Ae^t + Be^t) + 2(Ate^t + Be^t) = te^t $$ Simplifying, we get: $$ (-3A+2B)e^t = 1 $$ and $$ (-3A+4A)e^t = t $$ Comparing the coefficients of \(e^t\) and \(te^t\), we get: $$ A = 1,\quad B = \frac{1}{2} $$ So the particular integral (PI) is: $$ y_p(x) = te^t + \frac{1}{2}e^t $$

Obtain the General Solution

The general solution of the given differential equation can be obtained by adding the complementary function (CF) and the particular integral (PI): $$ y(x) = y_c(x) + y_p(x) = C_1 e^{2x} + C_2 e^{\frac{1}{2}x} + te^t + \frac{1}{2}e^t $$ This is the general solution of the given differential equation.

Key concepts

Homogeneous Equation

A homogeneous equation in the context of differential equations is one where all terms are dependent on the function and its derivatives, and the equation equals zero. In our exercise, the original differential equation is \[ 2y'' - 5y' + 2y = te^t \]To find the homogeneous equation, we eliminate the non-homogeneous part, here represented by the term \( te^t \).
This results in the equation:\[ 2y'' - 5y' + 2y = 0 \]
This homogeneous equation allows us to focus solely on the behavior of the function and its derivatives without external forces or influences. It provides the groundwork for finding the complementary function, which is a critical step in solving differential equations.

Characteristic Equation

The characteristic equation is derived by examining the homogeneous equation. For linear constant coefficient differential equations, it allows us to find the roots that will help determine the complementary function. In our exercise, the homogeneous equation is:\[ 2y'' - 5y' + 2y = 0 \]The characteristic equation is obtained by assuming solutions of the form \( y = e^{rx} \), and substituting into the homogeneous equation, which results in the polynomial equation:\[ 2r^2 - 5r + 2 = 0 \]
This is a quadratic equation, and solving it using factoring yields the roots:\( r_1 = 2 \) and \( r_2 = \frac{1}{2} \).
These roots reflect the exponential form of the components of the complementary function.

Particular Integral

The particular integral (PI) is a particular solution to the non-homogeneous differential equation. It accounts for the non-zero right side of the original equation that isn't captured by the complementary function.In this exercise, the non-homogeneous term is \( te^t \).
To determine the particular integral, we use the form:\[ y_p(x) = At e^t + B e^t \]
Substituting this guess and its derivatives into the non-homogeneous equation allows us to equate coefficients of like terms. This process results in identifying suitable values for parameters \( A \) and \( B \).
Specifically, after substitution and simplification, we compare coefficients to find:

• \( A = 1 \)
• \( B = \frac{1}{2} \)

Thus, the particular integral is:\[ y_p(x) = te^t + \frac{1}{2}e^t \]

Undetermined Coefficients

The method of undetermined coefficients is an approach used to find a particular solution of non-homogeneous linear differential equations. It assumes a form for the particular integral that includes arbitrary coefficients, which are then determined through substitution into the differential equation.
The choice of form is guided by the non-homogeneous term. In our exercise:\[ te^t \]
suggests using a trial solution of:\[ y_p(x) = At e^t + B e^t \]
We then find the derivatives of this trial solution and substitute them back into the original differential equation.Through this substitution, we equate coefficients of the functions on both sides of the equation to find values for the unknown parameters. This method is especially useful when the non-homogeneous term is a simple function like polynomials, exponentials, or sine/cosine, allowing us to solve otherwise complex differential equations more straightforwardly.