Question

Use the method of variation of parameters to find a particular solution of the given differential equation. Then check your answer by using the method of undetermined coefficients. $$ 4 y^{\prime \prime}-4 y^{\prime}+y=16 e^{t / 2} $$

Short answer

#Question# Use the method of variation of parameters to find a particular solution to the following differential equation: $$ 4y''-4y'+y=16e^{t/2} $$ Verify your answer using the method of undetermined coefficients. #Answer# Using the method of variation of parameters, the particular solution to the given differential equation is: $$ y_p(t)=16te^{t/2} $$ Using the method of undetermined coefficients, we verified that the particular solution is indeed: $$ y_p(t)=16te^{t/2} $$

Step-by-step solution

Complementary solution

First, we need to find the complementary solution based on the homogeneous equation. The homogeneous equation is: $$ 4y''-4y'+y=0 $$ The characteristic equation for this equation is: $$ 4r^2 - 4r + 1 = 0 $$ Using the quadratic formula, we find the roots: $$ r_{1,2}=\frac{4 \pm \sqrt{16-16}}{8}=\frac{1}{2} $$ Since both roots are the same, the complementary solution is in the form: $$ y_c(t)=c_1e^{rt}+c_2te^{rt}=c_1e^{t/2}+c_2te^{t/2} $$

Variation of Parameters (Wronskian, Particular Solution)

We will use the method of variation of parameters to find a particular solution. Let \(y_1(t)=e^{t/2}\) and \(y_2(t)=te^{t/2}\). We need to find the Wronskian of these two functions: $$ W(t)=\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \\ \end{vmatrix} =\begin{vmatrix} e^{t/2} & te^{t/2} \\ \frac{1}{2}e^{t/2} & e^{t/2}+\frac{1}{2}te^{t/2} \\ \end{vmatrix} =\frac{1}{2}e^{t}\left(2-\frac{1}{2}t\right) $$ Now, we can find the particular solution using the formula: $$ y_p(t)=-y_1(t)\int\frac{y_2}{W}g(t)dt+y_2(t)\int\frac{y_1}{W}g(t)dt $$ Here, \(g(t)=16e^{t/2}\). So, we can rewrite the formula for \(y_p(t)\): $$ y_p(t)= -e^{t/2}\int\frac{te^{t/2}}{\frac{1}{2}e^{t}(2-\frac{1}{2}t)}\times 16e^{t/2}dt +te^{t/2}\int\frac{e^{t/2}}{\frac{1}{2}e^{t}(2-\frac{1}{2}t)}\times 16e^{t/2}dt $$ Calculate the integrals: $$ y_p(t)=-e^{t/2}\int 32\frac{1}{2-\frac{1}{2}t}dt+te^{t/2}\int 32\left(2-\frac{1}{2}t\right)^{-1}dt $$ $$ y_p(t)=-e^{t/2}\left[-64\ln\left(2-\frac{1}{2}t\right)\right]+e^{t/2}\left[64t\ln\left(2-\frac{1}{2}t\right)-128t\right] $$

Method of Undetermined Coefficients (Verify)

Now, we will verify our answer using the method of undetermined coefficients. First, assume a particular solution of the form: $$ y_p(t)=Ate^{t/2} $$ Differentiate \(y_p(t)\) one time: $$ y_p'(t)=A\frac{1}{2}e^{t/2}+Ate^{\frac{t}{2}} $$ Differentiate \(y_p(t)\) two times: $$ y_p''(t)=A\frac{1}{4}e^{t/2}+A\frac{1}{2}e^{t/2}+Ate^{\frac{t}{2}} $$ Substitute \(y_p(t)\), \(y_p'(t)\), and \(y_p''(t)\) into the given differential equation: $$ 4(A\frac{1}{4}e^{t/2}+A\frac{1}{2}e^{t/2}+Ate^{\frac{t}{2}})-4(A\frac{1}{2}e^{t/2}+Ate^{\frac{t}{2}})+Ate^{t/2}=16e^{t/2} $$ Simplify and solve for A: $$ Ate^{t/2}=16e^{t/2} $$ $$ A=16 $$ So the particular solution using the method of undetermined coefficients is: $$ y_p(t)=16te^{t/2} $$ Comparing the results from both methods, we can see that the undetermined coefficients method gives a more simplified version of the particular solution, which is also included in the more general form found in step 2. So, the particular solution is verified as correct.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives of one or more dependent variables with respect to one or more independent variables. They are essential in expressing many natural phenomena, from basic kinematics in physics to complicated dynamics in engineering. A differential equation can often be categorized as either ordinary (ODE), involving derivatives with respect to a single variable, or partial (PDE), involving derivatives with respect to multiple variables. In essence, solving a differential equation means finding a function that satisfies the given relationship between its derivatives.

Wronskian Determinant

The Wronskian determinant is a crucial concept when dealing with second-order differential equations, particularly when employing the variation of parameters method to find particular solutions. Named after Hoene Wronski, the Wronskian is a determinant of a square matrix that contains the functions in question and their derivatives, organized in rows.

The value of the Wronskian tells us about the linear independence of the functions involved. If the determinant is non-zero on the interval of interest, the functions are linearly independent, which is a requisite for the method of variation of parameters to work effectively.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to find particular solutions to non-homogeneous ordinary differential equations. It is best used when the non-homogeneous part of the equation, often called the forcing function, is a relatively simple function, such as polynomials, exponentials, sines, and cosines. The method involves guessing a form for the particular solution where the coefficients are undetermined. These coefficients are later determined by plugging the guessed solution back into the differential equation and solving the resulting algebraic equations.

Characteristic Equation

Associated with linear homogeneous differential equations, the characteristic equation is an algebraic equation obtained by substituting a trial solution, usually in the form of an exponential function, into the differential equation. This trial solution typically transforms a differential equation into an algebraic one. By finding the roots of the characteristic equation, we can determine the complementary (or homogeneous) solution to the differential equation. The nature of the roots, whether they are real, complex, distinct, or repeated, will determine the form of the solution.

Particular Solution

In the context of differential equations, a particular solution is a single function that satisfies the non-homogeneous differential equation on a given domain. It is 'particular' because it includes specific values for any constants of integration. Finding the particular solution is a critical part of solving non-homogeneous equations, as the general solution is a sum of the complementary solution and the particular solution. There are several methods to find particular solutions, like the method of undetermined coefficients or the variation of parameters.

Complementary Solution

The complementary solution or homogeneous solution, denoted typically as y_c, refers to the general solution of the associated homogeneous differential equation (one with zero on the right side). It represents all possible solutions that satisfy this homogeneous equation. When solving a non-homogeneous differential equation, the complementary solution combines with a particular solution to form the complete solution to the equation.

Homogeneous Equation

A homogeneous differential equation is one in which every term is a function of the dependent variable and its derivatives.

More specifically, homogeneous refers to the fact that there is no standalone term or 'forcing term'; in other words, the equation is set to zero. Homogeneous differential equations are significant because their solutions can be used to construct the general solution of corresponding non-homogeneous equations through superposition, especially when the linearity principle applies.