Problem 15 - Advanced Engineering Mathematics

Question

Someone tells you that the solutions of the two IVPs $$ \begin{array}{ll} y^{\prime \prime}+2 y^{\prime}+10 y=0, & y(0)=0, y^{\prime}(0)=1 \\ \text { and } \quad y^{\prime \prime}+2 y^{\prime}+10 y=\delta(t), \quad y(0)=0, y^{\prime}(0)=0 \end{array} $$ are exactly the same. Do you agree or disagree? Defend your answer.

Step-by-Step Solution

Answer

Disagree, solutions differ by a factor \( 1/3 \).

Step 1: Analyze First IVP

For the homogeneous differential equation \( y'' + 2y' + 10y = 0 \) with initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \), we need to find its solution. The characteristic equation is \( r^2 + 2r + 10 = 0 \). Solving for \( r \) gives \( r = -1 \pm 3i \). Thus, the general solution is \( y_h(t) = e^{-t}(C_1 \cos(3t) + C_2 \sin(3t)) \). Applying initial conditions, solve for \( C_1 \) and \( C_2 \).

Step 2: Solve Homogeneous Solution

Applying the initial condition \( y(0) = 0 \), we substitute into the general solution to get \( y_h(0) = C_1 = 0 \). For the second initial condition \( y'(0) = 1 \), compute \( y'_h(t) = e^{-t}(-3C_1 \sin(3t) + 3C_2 \cos(3t)) - e^{-t}(C_1 \cos(3t) + C_2 \sin(3t)) \) and substitute \( t = 0 \) to find \( y'_h(0) = 3C_2 = 1 \), leading to \( C_2 = \frac{1}{3} \).

Step 3: Solution to First IVP

With \( C_1 = 0 \) and \( C_2 = \frac{1}{3} \), the solution to the first IVP is \( y_h(t) = \frac{1}{3} e^{-t} \sin(3t) \).

Step 4: Analyze Second IVP

For the inhomogeneous equation \( y'' + 2y' + 10y = \delta(t) \) with initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \), apply the Laplace transform to handle the delta function. Applying the Laplace transform, we have \( s^2 Y(s) + 2sY(s) + 10Y(s) = \mathcal{L}\{\delta(t)\} = 1 \).

Step 5: Solve with Laplace Transform

Solve for \( Y(s) \) using the Laplace transform: \( Y(s) = \frac{1}{(s^2 + 2s + 10)} \). Use partial fraction decomposition if necessary, and apply the inverse Laplace transform to determine \( y(t) \).

Step 6: Solution to Second IVP

After applying inverse Laplace transform, the particular solution and homogeneous solution combine to satisfy initial conditions. The result is \( y(t) = e^{-t} \sin(3t) \).

Step 7: Compare Both Solutions

The solution for the first IVP is \( \frac{1}{3} e^{-t} \sin(3t) \), whereas the solution for the second IVP, after applying the inverse Laplace transform, we find is \( e^{-t} \sin(3t) \). The factor \( \frac{1}{3} \) in the first solution indicates they're not identical.

Key Concepts

Initial Value Problem (IVP)

An Initial Value Problem, or IVP, involves solving differential equations with given conditions at a specific point, often where time is zero. This specific point is denoted as the initial condition. An IVP typically consists of a differential equation alongside one or several initial conditions. These initial conditions provide values for the function and its derivatives at a particular point. This is critical because it allows us to determine the specific solution that fits the situation.

In practical terms, you might encounter an IVP when modeling dynamic systems like the cooling of a hot object, the charge in a capacitor, or even the oscillation of a spring.

• Differential Equation: The main equation we aim to solve.
• Initial Conditions: Known values at the initial point(s).

For instance, in the first exercise, the initial conditions are: \( y(0) = 0 \) and \( y'(0) = 1 \). These help guide us to a precise solution rather than a general one. By inserting these into our solution form, we can adjust the constants to find the solution that meets these conditions.

Laplace Transform

The Laplace Transform is an invaluable tool for solving differential equations, specifically for transforming complicated functions into a simpler algebraic form that is easier to manipulate and solve. By converting a time-domain function into the s-domain, one can handle differential equations more as algebraic equations, which is usually far simpler.

When faced with an inhomogeneous differential equation that includes a delta function, like the second IVP in the exercise, the Laplace Transform is particularly useful. The delta function, \( \delta(t) \), simulates an instant impulse at a specific point in time, and its transform is 1.

• Transform special functions like the step function and delta function easily.
• Instantly manage linearity of differential operators.

To solve the second IVP using the Laplace Transform, you start by taking the transform of both sides of the equation. This converts derivatives into multiplication operations, simplifying the problem significantly. Once in the Laplace domain, you solve for the transformed function and then use the inverse Laplace Transform to convert back to the time domain and find the actual solution, as shown in the provided exercise.

Homogeneous Differential Equation

A Homogeneous Differential Equation is one that has zero on the right-hand side of the equation. In essence, all terms depend only on the function and its derivatives, without any "external" forces adding non-zero factors.

In our discussion, the equation \( y'' + 2y' + 10y = 0 \) is homogeneous because all terms are about the unknown function \( y \) and its derivatives, without other explicit functions of the independent variable (like a delta function or a sine wave) involved.

• Solve using characteristic equations to find natural frequencies of systems.
• Deals with modeling systems where output depends solely on internal factors.

To find the solution of a homogeneous differential equation like the one in the given exercise, we first formulate the characteristic equation (a quadratic equation derived from replacing derivatives with powers of \( r \)), find its roots, and then build the solution from these roots. This usually leads to solving for constants that fit initial conditions, which describe the specific scenario being analyzed.