Question

In Exercises \(1-20\) solve the initial value problem. Where indicated by \(\mathrm{C} / \mathrm{G}\), graph the solution. $$ 2 y^{\prime \prime}-3 y^{\prime}-2 y=1+\delta(t-2), \quad y(0)=-1, \quad y^{\prime}(0)=2 $$

Short answer

The final solution for the given initial value problem is: \(y(t) = \frac{3}{2} e^{2t} - 2 e^{-\frac{1}{2}t} - \frac{1}{2} + u(t-2)\), where \(u(t-2)\) is the unit step function.

Step-by-step solution

Solve the homogeneous equation

The homogeneous equation associated with the given differential equation is: $$ 2y'' - 3y' - 2y = 0 $$ To solve this, we have to find the characteristic equation, which is: $$ 2r^2 - 3r - 2 = 0 $$ Now, we'll solve for \(r\).

Find the roots of the characteristic equation

To find the roots of the characteristic equation, we can use the quadratic formula: $$ r = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$ Plugging in the values \(a=2\), \(b=-3\), and \(c=-2\) gives us: $$ r = \frac{3 \pm \sqrt{(-3)^2 -4(2)(-2)}}{4} $$ Solving this expression gives us \(r_1=2\) and \(r_2=-\frac{1}{2}\).

Write down the general solution for the homogeneous equation

Since we have two distinct roots, the general solution for the homogeneous equation is given by: $$ y_h(t) = C_1 e^{2t} + C_2 e^{-\frac{1}{2}t} $$ We will later determine the values of \(C_1\) and \(C_2\) using the given initial conditions.

Find the particular solution for the external force

Now, let's look for a particular solution for the given external force \(1+\delta(t-2)\). We can split the external force into two parts: a constant part and an impulse part, and find a particular solution for each part separately: $$ y_p(t) = y_{p1}(t) + y_{p2}(t) $$ For the constant part \((1)\), we can assume a particular solution of the form \(y_{p1}(t) = A\), where \(A\) is a constant. Plugging this into the given differential equation, we get: $$ 2(0) - 3(0) - 2A = 1 $$ Solving for \(A\), we get \(A=-\frac{1}{2}\), so \(y_{p1}(t)=-\frac{1}{2}\). For the impulse part \((\delta(t-2))\), we can assume a particular solution of the form \(y_{p2}(t) = B \cdot u(t-2)\), where \(B\) is a constant and \(u(t-2)\) is the unit step function. To find \(B\), we need to integrate the external force from \(2^-\) to \(2^+\): $$ \int_{2^-}^{2^+} 1 + \delta(t-2) dt = B $$ Evaluating the integral, we find that \(B=1\), so \(y_{p2}(t)=u(t-2)\). Thus, the particular solution for the given external force is \(y_p(t) = -\frac{1}{2} + u(t-2)\).

Combine the general and particular solutions

Now we can add the general solution for the homogeneous equation \(y_h(t)\) and the particular solution for the external force \(y_p(t)\) to form the complete solution for the given differential equation: $$ y(t) = y_h(t) + y_p(t) = C_1 e^{2t} + C_2 e^{-\frac{1}{2}t} - \frac{1}{2} + u(t-2) $$

Use the initial conditions to determine constants

We are given two initial conditions: \(y(0)=-1\) and \(y'(0)=2\). Using these, we can determine the constants \(C_1\) and \(C_2\). First, let's plug in \(t=0\) and find \(C_1\) and \(C_2\): $$ -1 = C_1 e^{0} + C_2 e^{0} - \frac{1}{2} + u(0-2) $$ Solving for \(C_1\) and \(C_2\), we get: $$ C_1 + C_2 = -\frac{1}{2} $$ Now, let's find the derivative of the complete solution: $$ y'(t) = 2C_1 e^{2t} -\frac{1}{2}C_2 e^{-\frac{1}{2}t} + \delta(t-2) $$ Plugging in \(t=0\), we get: $$ 2 = 2C_1 e^{0} -\frac{1}{2}C_2 e^{0} + \delta(0-2) $$ Solving for \(C_1\) and \(C_2\), we get: $$ 2C_1 -\frac{1}{2}C_2 = 2 $$ Solving the system of equations, we find \(C_1 = \frac{3}{2}\) and \(C_2 = -2\).

Write down the final solution

Substituting the values of \(C_1\) and \(C_2\) back into the complete solution, we get the final solution for the given initial value problem: $$ y(t) = \frac{3}{2} e^{2t} - 2 e^{-\frac{1}{2}t} - \frac{1}{2} + u(t-2) $$

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) in differential equations is about finding a solution to a differential equation that satisfies certain conditions at the starting point, usually provided at a specific value of the independent variable, often time. This is essential in many physical applications where initial conditions set the specific trajectory or behavior of a system.

In our exercise, we have the differential equation: \(2 y'' - 3 y' - 2 y = 1 + \delta(t-2)\), with initial conditions: \(y(0)=-1\) and \(y'(0)=2\). These conditions tell us the state of the system at time \(t=0\). This means that any solution we find must pass through this point, fitting these exact conditions at that initial moment.

Homogeneous Equation

A homogeneous differential equation is a type where all terms depend on the function or its derivatives and equals zero. These are significant because they represent the system's natural behavior, without external influences.

In the given exercise, the homogeneous part is \(2y'' - 3y' - 2y = 0\). Solving this gives the natural response of the system, describing how it behaves without any external forces or additions. The solution involves finding the roots of the associated characteristic equation, providing us with the solutions in terms of exponential functions, which form the basis of the system's intrinsic behavior.

Particular Solution

The particular solution in a differential equation problem deals with the non-homogeneous aspects – those caused by external inputs or forces. It represents the forced response of the system, as opposed to its natural behavior.

In this problem, the external force is shown by \(1 + \delta(t-2)\). We solve for the particular solution by addressing each external component separately: the constant force and the impulse function \(\delta(t-2)\). This leads to solutions like \(-\frac{1}{2}\) and \(u(t-2)\), describing how the system specifically reacts to these influences.

Characteristic Equation

A characteristic equation arises when solving linear homogeneous differential equations. It emerges from substituting exponential assumptions into the differential equation, leading to an algebraic equation. Solving this algebraic equation yields the roots critical to defining the system's natural response.

In our case, the characteristic equation \(2r^2 - 3r - 2 = 0\) is solved to find the roots \(r_1 = 2\) and \(r_2 = -\frac{1}{2}\). These roots delineate the nature of the solution to the homogeneous equation, forming exponentials that are either growing or decaying, thus providing a comprehensive view of the system’s behavior.