Question

(a) Find the general solution of the differential equation. (b) Impose the initial conditions to obtain the unique solution of the initial value problem. (c) Describe the behavior of the solution \(y(t)\) as \(t \rightarrow-\infty\) and as \(t \rightarrow \infty\). Does \(y(t)\) approach \(-\infty,+\infty\), or a finite limit? $$y^{\prime \prime}-5 y^{\prime}+6 y=0, \quad y(0)=1, \quad y^{\prime}(0)=-1$$

Short answer

Answer: As \(t \rightarrow -\infty\), \(y(t)\) approaches 0, and as \(t \rightarrow \infty\), \(y(t)\) approaches \(+\infty\).

Step-by-step solution

Find the general solution of the differential equation

To solve the given differential equation, we first need to determine the characteristic equation of the equation. The characteristic equation is given by: $$r^2 - 5r + 6 = 0$$ Next, we solve this equation for the roots (r).

Solve the characteristic equation

To find the roots of the characteristic equation, we can use the quadratic formula: $$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ In our case, \(a = 1\), \(b = -5\), and \(c = 6\). Plugging these values into the formula, we get: $$r = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)}$$ $$r = \frac{5 \pm \sqrt{25 - 24}}{2}$$ $$r = \frac{5 \pm 1}{2}$$ This gives us two roots, \(r_1 = 2\) and \(r_2 = 3\). With these roots, we can form the general solution of the differential equation as: $$y(t) = C_1e^{2t} + C_2e^{3t}$$

Use the initial conditions to find the unique solution

We have the initial conditions \(y(0) = 1\) and \(y'(0) = -1\). Using these conditions, we will find the values of \(C_1\) and \(C_2\). For the first initial condition: $$y(0) = C_1e^{2(0)} + C_2e^{3(0)}$$ $$1 = C_1 + C_2$$ Now we need to find \(y'(t)\). To do this, we differentiate \(y(t)\) with respect to \(t\): $$y'(t) = 2C_1e^{2t} + 3C_2e^{3t}$$ Now we use the second initial condition \(y'(0) = -1\): $$-1 = 2C_1e^{2(0)} + 3C_2e^{3(0)}$$ $$-1 = 2C_1 + 3C_2$$ We now have a system of linear equations: $$C_1 + C_2 = 1$$ $$2C_1 + 3C_2 = -1$$

Solve the system of linear equations

To find the values of \(C_1\) and \(C_2\), we can use the substitution or elimination method. We will use the elimination method: Multiply the first equation by 2 and subtract the second equation. This eliminates \(C_1\): $$2(C_1 + C_2) - (2C_1 + 3C_2) = 2(1) - (-1)$$ $$C_2 = 3$$ Now substitute the value of \(C_2\) in the first equation to find the value of \(C_1\): $$C_1 + 3 = 1$$ $$C_1 = -2$$ So, the unique solution of the initial value problem is: $$y(t) = -2e^{2t} + 3e^{3t}$$

Analyze the behavior of the solution

To find the behavior of the solution \(y(t)\) as \(t \rightarrow -\infty\) and \(t \rightarrow \infty\), we need to analyze the terms \(-2e^{2t}\) and \(3e^{3t}\). As \(t \rightarrow -\infty\), both terms approach 0, and thus \(y(t) \rightarrow 0\). As \(t \rightarrow \infty\), both terms approach \(\infty\), but \(3e^{3t}\) grows faster than \(-2e^{2t}\), so the entire expression approaches \(+\infty\). Therefore, as \(t \rightarrow -\infty\), \(y(t)\) approaches 0, and as \(t \rightarrow \infty\), \(y(t)\) approaches \(+\infty\).

Key concepts

General Solution

A general solution of a differential equation provides a family of functions that satisfy the equation. It captures the various possible solutions by incorporating arbitrary constants. In the given exercise, we start with the second-order linear differential equation:\[ y'' - 5y' + 6y = 0. \]To find its general solution, we need to determine the characteristic equation associated with the differential equation. This is a key step in the process, as it involves replacing each derivative in the equation with terms of a polynomial in terms of a variable \( r \) to reflect exponential solutions.The characteristic equation for the given differential equation is:\[ r^2 - 5r + 6 = 0. \]By solving this quadratic equation, we find the roots \( r_1 = 2 \) and \( r_2 = 3 \). These roots indicate that the general solution to our differential equation is a combination of exponential functions corresponding to those roots, given as:\[ y(t) = C_1 e^{2t} + C_2 e^{3t}, \]where \( C_1 \) and \( C_2 \) are arbitrary constants that can be determined if initial conditions are provided. This expression represents the general solution, accommodating all potential solutions based on different initial parameters.

Initial Conditions

Initial conditions are specifications at a particular point that help pinpoint the unique solution within the general solution's family of functions. These conditions are crucial for solving what's known as an initial value problem.In the presented exercise, we have the initial conditions: - \( y(0) = 1 \) - \( y'(0) = -1 \)These conditions allow us to determine the exact constants \( C_1 \) and \( C_2 \) in the general solution \( y(t) = C_1 e^{2t} + C_2 e^{3t} \). Substituting \( t = 0 \) in the general solution gives:- \( 1 = C_1 + C_2 \)For the derivative of \( y \), computing \( y'(t) = 2C_1e^{2t} + 3C_2e^{3t} \), and setting \( t = 0 \), we find:- \( -1 = 2C_1 + 3C_2 \)We now solve these linear equations to find \( C_1 \) and \( C_2 \):

• Substitute \( C_1 = 1 - C_2 \) into the equation \( 2C_1 + 3C_2 = -1 \).
• Solve to get \( C_2 = 3 \) and substituting back gives \( C_1 = -2 \).

Thus, the unique solution satisfying the initial conditions is:\[ y(t) = -2e^{2t} + 3e^{3t}. \]

Characteristic Equation

The characteristic equation is a fundamental concept when solving linear differential equations with constant coefficients. This mathematical tool allows us to transform a differential equation into an algebraic equation that can be more easily solved.For the differential equation \( y'' - 5y' + 6y = 0 \), the characteristic equation is obtained by substituting \( y(t) = e^{rt} \) into the differential equation, assuming a solution of exponential form:\[ r^2 - 5r + 6 = 0. \]This is a quadratic equation, and solving it gives the roots \( r_1 = 2 \) and \( r_2 = 3 \). These roots provide crucial information:

• They determine the form of the exponentials in the general solution.
• Each root corresponds to an exponential function \( e^{2t} \) and \( e^{3t} \), respectively.

For quadratic equations, depending on the nature of the roots - real and distinct, real and repeated, or complex conjugates - the solution structure might change. Here, since the roots are real and distinct, our general solution is a simple linear combination of two exponential terms:\[ y(t) = C_1 e^{2t} + C_2 e^{3t}. \]This characteristic equation not only helps in finding generalized solutions but also informs us as to how these solutions behave. For instance, positive roots indicate exponentially growing solutions. It's a key analytical step to understanding the dynamics of differential equations.