Question

Find a fundamental set \(\left\\{\bar{y}_{1}, \bar{y}_{2}\right\\}\) satisfying the given initial conditions. \(y^{\prime \prime}+4 y^{\prime}+5 y=0, \quad \bar{y}_{1}(0)=1, \quad \bar{y}_{1}^{\prime}(0)=-1, \quad \bar{y}_{2}(0)=0, \quad \bar{y}_{2}^{\prime}(0)=1\)

Short answer

Question: Determine the fundamental set of solutions for the given homogeneous second-order linear differential equation, \(y^{\prime\prime}+4y^{\prime}+5y=0\), with the initial conditions \(\bar{y}_1(0) = 1\), \(\bar{y}_1^{\prime}(0) = -1\), \(\bar{y}_2(0) = 0\), and \(\bar{y}_2^{\prime}(0) = 1\). Answer: The fundamental set of solutions is \(\left\\{\bar{y}_{1}, \bar{y}_{2}\right\\} = \left\\{e^{-2x}(\cos{x} - \sin{x}), e^{-2x}\sin{x}\right\\}\).

Step-by-step solution

Find the characteristic equation

To solve the given homogeneous linear differential equation, \(y^{\prime\prime}+4y^{\prime}+5y=0\), we first need to find the characteristic equation, which is obtained by assuming a solution of the form \(y = e^{rx}\). The characteristic equation will have the form \(r^2 + 4r + 5 = 0\).

Solve the characteristic equation

Now, we will solve the quadratic equation, \(r^2 + 4r + 5 = 0\). Since the discriminant (\(\Delta = 4^2 - 4 \cdot 5 < 0\)), the roots are complex conjugates. The roots can be found using the quadratic formula: \(r = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-4 \pm \sqrt{-4}}{2} = -2 \pm i\).

Find the general solution

Given that the roots are complex conjugates, the general solution to the homogeneous linear differential equation will be in the form: \(y(x) = e^{-2x}(c_1 \cos{x} + c_2 \sin{x})\).

Satisfy the initial conditions

We are given the initial conditions \(\bar{y}_1(0) = 1\), \(\bar{y}_1^{\prime}(0) = -1\), \(\bar{y}_2(0) = 0\), and \(\bar{y}_2^{\prime}(0) = 1\). Using these initial conditions, we will now determine the constants \(c_1\) and \(c_2\) in the general solution for both \(\bar{y}_1\) and \(\bar{y}_2\). First, for \(\bar{y}_1(0) = 1\), plug in \(x=0\) to find: \(1 = e^{0}(c_1 \cos{0} + c_2 \sin{0}) \Rightarrow c_1 = 1\). Next, for \(\bar{y}_1^{\prime}(0) = -1\), take the derivative of \(\bar{y}_1(x)\) and plug in \(x=0\): \(\bar{y}_1^{\prime}(x) = e^{-2x}(-2c_1 \cos{x} - 2c_2 \sin{x} + c_2 \cos{x} - c_1 \sin{x})\). Plugging in \(x=0\), we get: \(-1 = e^{0}(-2 \cdot 1 \cos{0} - 2c_2 \sin{0} + c_2 \cos{0} - 1 \sin{0}) \Rightarrow c_2 = -1\). Thus, for \(\bar{y}_1(x)\), we have: \(\bar{y}_1(x) = e^{-2x}(\cos{x} - \sin{x})\). Now, for \(\bar{y}_2(0) = 0\), plug in \(x=0\) to find: \(0 = e^{0}(c_1 \cos{0} + c_2 \sin{0}) \Rightarrow c_1 = 0\). Finally, for \(\bar{y}_2^{\prime}(0) = 1\), take the derivative of \(\bar{y}_2(x)\) and plug in \(x=0\): \(\bar{y}_2^{\prime}(x) = e^{-2x}(-2c_1 \cos{x} - 2c_2 \sin{x} + c_2 \cos{x} - c_1 \sin{x})\). Plugging in \(x=0\), we get: \(1 = e^{0}(-2 \cdot 0 \cos{0} - 2c_2 \sin{0} + c_2 \cos{0} - 0 \sin{0}) \Rightarrow c_2 = 1\). Thus, for \(\bar{y}_2(x)\), we have: \(\bar{y}_2(x) = e^{-2x}(\sin{x})\).

Write the fundamental set

Now we have found the solutions \(\bar{y}_1(x)\) and \(\bar{y}_2(x)\), we can write the fundamental set: \(\left\\{\bar{y}_{1}, \bar{y}_{2}\right\\} = \left\\{e^{-2x}(\cos{x} - \sin{x}), e^{-2x}\sin{x}\right\\}\).

Key concepts

Characteristic Equation

Understanding the characteristic equation is crucial when dealing with homogeneous linear differential equations. It's a gateway to finding the general solution to the equation. Let's illustrate this with the equation given in the exercise, where we have to solve:
\[y^{\text{''}} + 4y^{\text{'}} + 5y = 0.\]
To get the characteristic equation, we assume a solution of the form \(y = e^{rx}\) and substitute it into our differential equation. Here, the characteristic equation takes the form:
\[r^2 + 4r + 5 = 0.\]
Solving this quadratic provides us with the roots or, in the case of a differential equation, the eigenvalues. These roots will tell us the nature of the solution to the differential equation — whether it's real or complex, single or repeated. For the characteristic equation above, because the discriminant is negative, we obtained complex conjugate roots, \(-2 \text{pm} i\), which led us to the solution involving exponential, sine, and cosine functions. Always remember, the characteristic equation is the backbone for finding the general solution to homogeneous differential equations.

Homogeneous Linear Differential Equation

The concept of a homogeneous linear differential equation refers to an equation that can be written in the form:
\[a_ny^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_1y^{\text{'}} + a_0y = 0,\]
where \(a_n, a_{n-1}, \ldots, a_1, a_0\) are constant coefficients, and \(y^{(n)}\) represents the nth derivative of \(y\) with respect to \(x\). The term 'homogeneous' implies that there is no additional function of \(x\) on the right side of the equation; it's set to zero. The solution to such an equation is greatly facilitated by understanding and applying the concept of the characteristic equation, as seen in Step 1 of our solution. This approach transforms the differential equation into an algebraic one whose solutions characterize the form of the general solution to the original differential equation.

Initial Conditions

The initial conditions are specific values given at a particular point, usually when \(x=0\), which allow us to find the unique solution to a differential equation from the family of general solutions. In our exercise, the initial conditions were:
\[\bar{y}_1(0) = 1, \; \bar{y}_1^{\text{'}}(0) = -1, \; \bar{y}_2(0) = 0, \; \bar{y}_2^{\text{'}}(0) = 1.\]
Applying these initial conditions helped us determine the unknown coefficients \(c_1\) and \(c_2\) in the general solution. It's akin to fitting the solution to satisfy the given starting criteria, making the solution applicable to the specific scenario posed by the problem. It's like having a key that perfectly fits a lock, primary to solving any differential equation problem that includes finding a particular solution.