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? $$2 y^{\prime \prime}-y=0, \quad y(0)=-2, \quad y^{\prime}(0)=\sqrt{2}$$

Short answer

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

Step-by-step solution

Find the general solution of the differential equation

To find the general solution of the equation, first, we need to find the characteristic equation of the given differential equation. The characteristic equation of the differential equation \(ay^{\prime \prime} + by^{\prime} + cy = 0\) is \(ar^2 + br + c = 0\) Here, the given differential equation is \(2y^{\prime \prime} - y = 0\). Therefore, the characteristic equation is: $$2r^2 - 1 = 0$$ To find the roots, we can use the quadratic formula: $$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, \(a = 2, b = 0, c = -1\) $$r = \frac{\pm \sqrt{4}}{4}$$ The roots of the characteristic equation are: $$r_1 = \frac{1}{\sqrt{2}}, \quad r_2 = -\frac{1}{\sqrt{2}}$$ Now we can write the general solution of the given differential equation: $$y(t) = c_1 e^{\frac{1}{\sqrt{2}}t} + c_2 e^{-\frac{1}{\sqrt{2}}t}$$

Impose the initial conditions to find the unique solution

Now, we need to find the derivatives of the general solution in order to impose the initial conditions: $$y^{\prime}(t) = \frac{1}{\sqrt{2}}c_1 e^{\frac{1}{\sqrt{2}}t} - \frac{1}{\sqrt{2}}c_2 e^{-\frac{1}{\sqrt{2}}t}$$ Initial conditions: \(y(0) = -2\) and \(y^{\prime}(0) = \sqrt{2}\) Apply the initial condition \(y(0) = -2\): $$-2 = c_1 e^{0} + c_2 e^{0} \Rightarrow -2 = c_1 + c_2$$ Apply the initial condition \(y^{\prime}(0) = \sqrt{2}\): $$\sqrt{2} = \frac{1}{\sqrt{2}}c_1 e^{0} - \frac{1}{\sqrt{2}}c_2 e^{0} \Rightarrow \sqrt{2} = \frac{1}{\sqrt{2}}(c_1 - c_2)$$ Solve the system of equations: $$c_1 = -\frac{3}{2}, \quad c_2 = \frac{1}{2}$$ Hence, the unique solution is: $$y(t) = -\frac{3}{2}e^{\frac{1}{\sqrt{2}}t} + \frac{1}{2}e^{-\frac{1}{\sqrt{2}}t}$$

Describe the behavior of the solution when \(t \rightarrow -\infty\) and \(t \rightarrow \infty\)

As \(t \rightarrow -\infty\), the terms in the solution behave as follows: $$e^{\frac{1}{\sqrt{2}}t} \rightarrow \infty$$ $$e^{-\frac{1}{\sqrt{2}}t} \rightarrow 0$$ Thus, \(y(t) \rightarrow -\infty\) as \(t \rightarrow -\infty\) As \(t \rightarrow \infty\), the terms in the solution behave as follows: $$e^{\frac{1}{\sqrt{2}}t} \rightarrow \infty$$ $$e^{-\frac{1}{\sqrt{2}}t} \rightarrow 0$$ Thus, \(y(t) \rightarrow \infty\) as \(t \rightarrow \infty\) In conclusion, the solution \(y(t)\) approaches \(-\infty\) as \(t \rightarrow -\infty\) and \(+\infty\) as \(t \rightarrow \infty\).

Key concepts

Characteristic Equation

In differential equations, the characteristic equation is crucial for solving linear homogeneous equations. It allows us to find the roots, which are key in building the general solution.

For a second-order linear homogeneous differential equation like \(2y'' - y = 0\), you need to find its characteristic equation.

The standard form is \(ay'' + by' + cy = 0\) and the characteristic equation is \(ar^2 + br + c = 0\). Here, \(a = 2\), \(b = 0\), and \(c = -1\), giving us \(2r^2 - 1 = 0\).

Solving this, we find the roots using the quadratic formula:

• The formula yields \(r = \frac{\pm\sqrt{4}}{4}\).
• The roots are \(r_1 = \frac{1}{\sqrt{2}}\) and \(r_2 = -\frac{1}{\sqrt{2}}\).

These roots guide us in constructing the general solution for the differential equation.

General Solution

The general solution of a differential equation involves a combination of exponential functions derived from the roots of the characteristic equation.

Once we have the roots \(r_1 = \frac{1}{\sqrt{2}}\) and \(r_2 = -\frac{1}{\sqrt{2}}\), the general solution can be expressed using these roots:

• \(y(t) = c_1 e^{\frac{1}{\sqrt{2}}t} + c_2 e^{-\frac{1}{\sqrt{2}}t}\).
• Here, \(c_1\) and \(c_2\) are arbitrary constants that will be determined by initial conditions.

This solution represents all possible functions \(y(t)\) satisfying the given differential equation, each corresponding to different values of \(c_1\) and \(c_2\).

Initial Value Problem

An initial value problem specifies the conditions that the solution of a differential equation must satisfy at a particular point. The aim is to find a unique solution within the family of possible solutions.

Given the initial conditions \(y(0) = -2\) and \(y'(0) = \sqrt{2}\), we substitute these into the general solution to find the specific constants \(c_1\) and \(c_2\):

• From \(y(0) = -2\), we get \(-2 = c_1 + c_2\).
• From \(y'(0) = \sqrt{2}\), we derive \(\sqrt{2} = \frac{1}{\sqrt{2}}(c_1 - c_2)\).

Solving these equations, we obtain the constants \(c_1 = -\frac{3}{2}\) and \(c_2 = \frac{1}{2}\).

Therefore, the unique solution that meets these conditions is:

• \(y(t) = -\frac{3}{2}e^{\frac{1}{\sqrt{2}}t} + \frac{1}{2}e^{-\frac{1}{\sqrt{2}}t}\).

Behavior of Solutions

Understanding the behavior of solutions is key to grasping what happens to \(y(t)\) as \(t\) changes. This involves examining what occurs as \(t\) approaches \(-\infty\) or \(\infty\).

Let's look at \(y(t) = -\frac{3}{2}e^{\frac{1}{\sqrt{2}}t} + \frac{1}{2}e^{-\frac{1}{\sqrt{2}}t}\):

• As \(t \rightarrow -\infty\): The term \(e^{\frac{1}{\sqrt{2}}t} \rightarrow 0\) and \(e^{-\frac{1}{\sqrt{2}}t} \rightarrow \infty\). This results in \(y(t) \rightarrow -\infty\).
• As \(t \rightarrow \infty\): \(e^{\frac{1}{\sqrt{2}}t} \rightarrow \infty\) and \(e^{-\frac{1}{\sqrt{2}}t} \rightarrow 0\), leading to \(y(t) \rightarrow \infty\).

This tells us that as time progresses, the solution diverges to both \(-\infty\) and \(+\infty\), depending on the direction of \(t\).