Question

Find a fundamental set \(\left\\{\bar{y}_{1}, \bar{y}_{2}\right\\}\) satisfying the given initial conditions. \(y^{\prime \prime}+4 y^{\prime}+4 y=0, \quad \bar{y}_{1}(0)=-1, \quad \bar{y}_{1}^{\prime}(0)=2, \quad \bar{y}_{2}(0)=1, \quad \bar{y}_{2}^{\prime}(0)=-1\)

Short answer

Question: Determine the fundamental set of solutions to the given second-order homogeneous linear differential equation \(y^{\prime \prime}+4 y^{\prime}+4 y=0\) satisfying the initial conditions: 1. \(\bar{y}_{1}(0)=-1\) and \(\bar{y}_{1}^{\prime}(0)=2\) 2. \(\bar{y}_{2}(0)=1\) and \(\bar{y}_{2}^{\prime}(0)=-1\) Answer: The fundamental set of solutions satisfying the given initial conditions is \(\left\\{\bar{y}_{1}, \bar{y}_{2}\right\\}=\left\\{-e^{-2x}, e^{-2x}+xe^{-2x} \right\\}\).

Step-by-step solution

Solve the homogeneous linear differential equation

Let's first solve the homogeneous linear differential equation \(y^{\prime \prime}+4 y^{\prime}+4 y=0\). The characteristic equation for this differential equation is given by: \(r^2+4r+4=0\) Let's solve for r to find the roots: \((r+2)^2=0\) \(r=-2\) Since we have a repeated root, the general solution for the homogeneous linear differential equation is given by: \(y(x)=c_{1}e^{-2x}+c_{2}xe^{-2x}\)

Apply the initial conditions to find \(\bar{y}_{1}(x)\) and \(\bar{y}_{2}(x)\)

Now, we need to find the specific functions \(\bar{y}_{1}(x)\) and \(\bar{y}_{2}(x)\) that satisfy the initial conditions. For \(\bar{y}_{1}(x)\), we have the following initial conditions: \(\bar{y}_{1}(0)=-1\) \(\bar{y}_{1}^{\prime}(0)=2\) Let's substitute the initial conditions in the general solution: \(\bar{y}_{1}(0)=-1=c_{1}e^0+c_{2}xe^0=-1\) \(c_{1}=-1\) Now, let's find the derivative of the general solution: \(\bar{y}^{\prime}(x)=-2c_{1}e^{-2x}+c_{2}e^{-2x}(1-2x)\) Substitute the initial condition for the derivative: \(\bar{y}_{1}^{\prime}(0)=2=-2c_{1}+c_{2}\) \(2=2+c_{2}\) \(c_{2}=0\) Thus, the function \(\bar{y}_{1}(x)\) is given by: \(\bar{y}_{1}(x)=-e^{-2x}\) For \(\bar{y}_{2}(x)\), we have the following initial conditions: \(\bar{y}_{2}(0)=1\) \(\bar{y}_{2}^{\prime}(0)=-1\) Substitute the initial conditions in the general solution: \(\bar{y}_{2}(0)=1=c_{1^{\prime}}e^0+c_{2^{\prime}}xe^0=1\) \(c_{1^{\prime}}=1\) Now, substitute the initial condition for the derivative: \(\bar{y}_{2}^{\prime}(0)=-1=-2c_{1^{\prime}}+c_{2^{\prime}}\) \(-1=-2+c_{2^{\prime}}\) \(c_{2^{\prime}}=1\) Thus, the function \(\bar{y}_{2}(x)\) is given by: \(\bar{y}_{2}(x)=e^{-2x}+xe^{-2x}\) Finally, the fundamental set satisfying the given initial conditions is \(\left\\{\bar{y}_{1}, \bar{y}_{2}\right\\}=\left\\{-e^{-2x}, e^{-2x}+xe^{-2x} \right\\}\).

Key concepts

Characteristic Equation

In solving homogeneous linear differential equations, the characteristic equation plays a crucial role. This equation helps us identify the roots necessary for forming the general solution. For the given differential equation \[ y'' + 4y' + 4y = 0, \]we derive the characteristic equation by substituting the derivatives with powers of \( r \), leading to \[ r^2 + 4r + 4 = 0. \]This is a standard quadratic equation where the solutions for \( r \) provide us with the roots required for constructing the solution to the differential equation. Remember, these roots dictate the behaviour of the solution, telling us if the system oscillates, decays, or grows over time.

Repeated Roots

Encountering repeated roots in the characteristic equation adds an interesting twist to finding solutions. In our example, the characteristic equation \[ r^2 + 4r + 4 = 0 \]can be factored into \[ (r + 2)^2 = 0, \]yielding the repeated root \( r = -2 \). Repeated roots mean that the associated solutions need to be modified slightly to form a complete fundamental set. Instead of basic exponential solutions, a repeated root leads to solutions like

• \( y_1(x) = e^{-2x} \)
• \( y_2(x) = xe^{-2x} \)

This form ensures linear independence and completeness of the solution set.

Fundamental Set

A fundamental set of solutions, based on their linear independence, provides a full description of solutions for second-order linear homogeneous differential equations. From the previous sections, we derived that the solutions corresponding to our repeated root are \( \{-e^{-2x}, e^{-2x} + xe^{-2x} \} \). These functions serve as building blocks to tailor solutions in real-world applications meeting specific requirements. Any linear combination of this fundamental set can satisfy different conditions applicable to the given differential equation.

Initial Conditions

Initial conditions are pivotal in determining particular solutions from the fundamental set. They offer specific values for the function and its derivatives at a given point. For example, with initial conditions

• \( \bar{y}_1(0) = -1 \)
• \( \bar{y}_1'(0) = 2 \)
• \( \bar{y}_2(0) = 1 \)
• \( \bar{y}_2'(0) = -1 \)

we adjust the linear combination coefficients to meet these requirements. This personalizes the general solution based on the scenario described by these initial conditions, precisely influencing our functions to align with given criteria. Consequently, \( \bar{y}_1(x) = -e^{-2x} \) and \( \bar{y}_2(x) = e^{-2x} + xe^{-2x} \) satisfy the stated initial conditions, tailoring the fundamental set accordingly.