Question

In each of Problems 13 through 18 find the solution of the given initial value problem. $$ y^{\prime \prime}+y^{\prime}-2 y=2 t, \quad y(0)=0, \quad y^{\prime}(0)=1 $$

Short answer

Answer: The specific solution is \(y(t) = 0.5e^t - t - 0.5\).

Step-by-step solution

Find the complementary function using the characteristic equation

First, we need to find the characteristic equation of the homogeneous ODE: $$ r^2 + r - 2 = 0 $$ This can be factored as: $$ (r + 2)(r - 1) = 0 $$ So, we got two real and distinct roots: r1 = -2 and r2 = 1. The complementary function is: $$ y_c(t) = C_1 e^{-2t} + C_2 e^t $$

Find a particular solution using undetermined coefficients

Since the non-homogeneous term is 2t, we will assume a particular solution in the form $$ y_p(t) = At + B $$ Taking the first and second derivatives of this, we have $$ y_p^{\prime}(t) = A, \quad y_p^{\prime\prime}(t) = 0 $$ Plug the assumed particular solution and its derivatives into the given non-homogeneous ODE: $$ 0 + A - 2(At + B) = 2t $$ Simplify and arrange the terms for coefficients of t and the constant term: $$ -2At = 2t, \quad A - 2B = 0 $$ Comparing coefficients, we get the following system of equations: $$ -2A = 2, \quad A - 2B = 0 $$ Solve this system, and we find A = -1 and B = -0.5. Thus, the particular solution is: $$ y_p(t) = -t - 0.5 $$

Form the general solution

The general solution is the sum of the complementary function and the particular solution: $$ y(t) = y_c(t) + y_p(t) = C_1 e^{-2t} + C_2 e^t - t - 0.5 $$

Use the initial conditions to obtain the specific solution

We are given the following initial conditions: $$ y(0) = 0, \quad y^{\prime}(0) = 1 $$ First, let's find \(y^{\prime}(t)\): $$ y^{\prime}(t) = -2 C_1 e^{-2t} + C_2 e^t - 1 $$ Apply the initial condition \(y(0) = 0\): $$ 0 = C_1 e^0 + C_2 e^0 - 0 - 0.5 $$ This equation simplifies to: $$ C_1 + C_2 - 0.5 = 0 $$ Apply the initial condition \(y^{\prime}(0) = 1\): $$ 1 = -2 C_1 e^0 + C_2 e^0 - 1 $$ This equation simplifies to: $$ -2 C_1 + C_2 = 2 $$ Solve the system of equations for C1 and C2: $$ C_1 + C_2 = 0.5, \quad -2 C_1 + C_2 = 2 $$ We find C1 = 0 and C2 = 0.5. Thus, the specific solution is: $$ y(t) = 0.5e^t - t - 0.5 $$ This is the solution to the given initial value problem.

Key concepts

Complementary Function

When solving differential equations, particularly second-order linear homogeneous equations with constant coefficients, the complementary function (CF), also known as the homogeneous solution, is crucial in forming the general solution. The CF is obtained by solving the equation as if the right-hand side were zero.

The solution to the corresponding characteristic equation provides us with the roots that are essential for constructing the complementary function. In our problem, after determining the roots (\r_1 = -2\text{ and }\r_2 = 1\text{), we use them to build the CF as a linear combination of exponentials based on these roots:\r}

\rCF: \r

\r\(y_c(t) = C_1 e^{-2t} + C_2 e^t\)

\r

\r

In this expression, \(C_1\) and \(C_2\) are arbitrary constants that will be determined later using initial conditions. This part of the solution will always satisfy the homogeneous differential equation irrespective of the initial conditions.

Particular Solution

The particular solution (PS) is a single function that satisfies the non-homogeneous differential equation. It needs to be found separately from the complementary function, and it depends on the form of the non-homogeneous term. By finding the PS, we address the specific influence of the non-homogeneous part of the equation on the overall solution.

In our exercise, the non-homogeneous term is a linear function of \(t\), specifically \(2t\). To find an appropriate PS, we guess a solution in a similar form and determine its coefficients by substituting back into the differential equation:\r}

\rAssumed PS: \r

\r\(y_p(t) = At + B\)

\r

\r

Upon substitution and comparing coefficients, we deduce that \(A = -1\) and \(B = -0.5\), resulting in a particular solution of \(y_p(t) = -t - 0.5\). This solution will only solve the non-homogeneous part of the differential equation.

Undetermined Coefficients

Undetermined coefficients is a method used to determine the particular solution to a non-homogeneous differential equation. It involves making an educated guess about the form of the particular solution based on the non-homogeneous term. The assumed solution usually mirrors the form of the term but includes undetermined coefficients that need to be solved for.

In our case, since the non-homogeneous term was a first-degree polynomial \(2t\), we assumed the form of the particular solution to be a polynomial of the same degree, \(y_p(t) = At + B\). The process includes differentiating this assumed solution, plugging it into the differential equation, and then solving for the coefficients by equating the coefficients of like terms on both sides of the equation. This straightforward approach gives us a systematic way to find the coefficients that will make our assumed solution valid.

Characteristic Equation

The characteristic equation is a tool we use to determine the roots that are required to construct the complementary function for a homogeneous linear differential equation with constant coefficients.

To formulate the characteristic equation from our original differential equation \(y^{\r\prime \r\prime}+y^{\r\prime}-2y=0\), we replace each derivative with a power of \(r\), where \(r\) signifies the root. This gives us a polynomial equation:\r}

\rCharacteristic Equation:\r

\r\(r^2 + r - 2 = 0\)

\r

\r

We then factor this equation to find its roots, which in this instance are \(r_1 = -2\) and \(r_2 = 1\). These roots guide us in constructing the complementary function, and they indicate the type of solution we can expect—real distinct roots lead to real and distinct exponential functions in the solution.