91Ó°ÊÓ

An Initial Value Problem (IVP) is a type of differential equation that comes with specific conditions called initial conditions.
These conditions allow us to find a unique solution to the differential equation.
In this exercise, the differential equation is given along with the initial conditions, which are values for the function and its derivative at a specific point. For the problem at hand, we're dealing with:

• The differential equation: \[y''(t) + 2y'(t) + 5y(t) = 4 e^{-t}\]
• The initial conditions: \(y(0) = 1\) and \(y'(0) = 1\)

These initial conditions are essential because they help in determining the specific constants in our general solution, thereby narrowing it down to a particular solution that fits the given problem. Without these conditions, we can only derive a general solution that could fit an unlimited number of situations.

A Homogeneous Equation is a natural part of solving linear differential equations.
In this context, a homogeneous equation means a differential equation set equal to zero.The homogeneous part of the given problem is: \[y''(t) + 2y'(t) + 5y(t) = 0\] This part is separated from the non-homogeneous part (the part involving \(4 e^{-t}\)).
By tackling this homogeneous equation separately, we can find the complementary or general solution. Its roots can help solve the entire differential equation.For our specific case, dealing with complex roots:

• If roots are real and distinct: Solutions often involve exponential functions with constant coefficients.
• If roots are complex: Solutions involve a combination of sine and cosine functions with exponential parts.

Finding the Particular Solution is crucial for solving non-homogeneous differential equations.
The particular solution is needed to address the non-zero side of the equation.In our exercise, the non-homogeneous term is \(4 e^{-t}\), suggesting that the function itself is part of our guess for the particular solution:

• We start by assuming: \(y_p(t) = Ae^{-t}\)
• Compute its derivatives and substitute back into the differential equation.

This straightforward process yields an equation where we can solve for \(A\), ensuring that the particular solution, \(y_p(t)\), satisfies the original equation.
In this scenario, we've found that \(A=1\), leading to the particular solution \(y_p(t) = e^{-t}\).

The Characteristic Equation is a fundamental tool for analyzing and solving linear differential equations.
It comes into play when dealing with homogeneous equations.For our exercise, after isolating the homogeneous part:

• Assume a solution form: \(y_h(t) = e^{rt}\)
• Substitute back into the homogeneous equation to form a polynomial equation, known as the characteristic equation.

Given the equation: \[r^2 + 2r + 5 = 0\] We solve it to find the roots using the quadratic formula: \[r = \frac{-2 \pm \sqrt{4 - 20}}{2}\] The complex roots \(-1 \pm 2i\) indicate that our general solution will involve exponential, sine, and cosine terms: \[y_h(t) = e^{-t}(C_1 \cos(2t) + C_2 \sin(2t))\] This step sets the stage for combining with the particular solution to form the final solution.