Question

Find the solution of the given initial value problem and plot its graph. How does the solution behave as \(t \rightarrow \infty ?\) $$ y^{\prime \prime \prime}-y^{\prime \prime}+y^{\prime}-y=0 ; \quad y(0)=2, \quad y^{\prime}(0)=-1, \quad y^{\prime \prime}(0)=-2 $$

Short answer

The specific solution of the given initial value problem is: $$ y(t) = e^t + (1+i) e^{-it} + (1-i) e^{it} $$ As \(t \rightarrow \infty\), the solution tends towards infinity due to the exponential growth of the real exponential term (\(e^t\)).

Step-by-step solution

Find the Characteristic Equation

The given differential equation is $$ y^{\prime\prime\prime}-y^{\prime\prime}+y^{\prime}-y=0 $$ To find the characteristic equation, replace \(y\) with \(r^n\) and its derivatives with their corresponding powers of \(r\). The characteristic equation is then given by: $$ r^3-r^2+r-1=0 $$

Solve the Characteristic Equation

To solve the characteristic equation, \((r^3-r^2+r-1=0)\), we attempt to factor it. We find that \((r-1)\) is a factor, resulting in the factored equation: $$ (r-1)(r^2+1)=0 $$ This factored equation gives us three roots: \(r_1=1\) and \(r_2,r_3 = \pm i\) (complex conjugate roots).

Construct the General Solution

With the roots found in Step 2, we can now construct the general solution for the given differential equation. Since we have one real root and two complex conjugate roots, the general solution takes the form: $$ y(t) = c_1 e^t + c_2 e^{-it} + c_3 e^{it} $$ where \(c_1\), \(c_2\), and \(c_3\) are constants to be determined using the initial conditions.

Use the Initial Conditions to Find the Specific Solution

We now use the initial conditions \(y(0)=2\), \(y^{\prime}(0)=-1\), and \(y^{\prime\prime}(0)=-2\) to find the specific solution. First, take the first and second derivatives of the general solution: $$ y^{\prime}(t) = c_1 e^t - c_2 ie^{-it} + c_3 ie^{it} $$ $$ y^{\prime\prime}(t) = c_1 e^t + c_2 e^{-it} - c_3 e^{it} $$ Now, apply the initial conditions: $$ y(0) = c_1 + c_2 + c_3 = 2 $$ $$ y^{\prime}(0) = c_1 - c_2 i + c_3 i = -1 $$ $$ y^{\prime\prime}(0) = c_1 + c_2 - c_3 = -2 $$ Solving this system of equations, we find \(c_1=1\), \(c_2=1+i\), and \(c_3=1-i\). Therefore, the specific solution is: $$ y(t) = e^t + (1+i) e^{-it} + (1-i) e^{it} $$

Analyze the Behavior of the Solution as \(t \rightarrow \infty\)

To analyze the behavior of the solution as \(t \rightarrow \infty\), we consider the individual terms in the solution. The exponential term involving complex exponentials converges to zero as \(t \rightarrow \infty\), and the real exponential \((e^t)\) grows without bound as \(t \rightarrow \infty\). Therefore, the solution tends towards infinity as time grows unbounded.

Plot the Graph of the Solution

Use graphing software or a graphing calculator to plot the graph of the specific solution found in Step 4: $$ y(t) = e^t + (1+i) e^{-it} + (1-i) e^{it} $$ Note that the real part of the solution is sufficient for plotting the graph, as the imaginary parts cancel out. The graph will show an exponential growth as \(t \rightarrow \infty\).

Key concepts

Characteristic Equation

The characteristic equation is a critical step when solving linear differential equations. It involves finding a polynomial whose roots are used to construct the general solution to the differential equation. In our case, the differential equation is given as

• \(y^{\prime\prime\prime} - y^{\prime\prime} + y^{\prime} - y = 0\)

To derive the characteristic equation, we substitute each derivative term with powers of \(r\), treating \(y\) as \(\text{e}^{rt}\) where \(r\) represents the unknown roots. This substitution generates the polynomial equation:

• \(r^3 - r^2 + r - 1 = 0\)

By solving this equation, you'll find the roots necessary to form the general solution of the differential equation.

Complex Conjugate Roots

Complex conjugate roots often appear when solving polynomial equations in differential equations. Once you've factored the characteristic equation and found the roots, you might encounter complex numbers. Here, for the characteristic equation \(r^3 - r^2 + r - 1 = 0\), we have:

• \(r_1 = 1\)
• \(r_2, r_3 = \pm i\)

The complex conjugate roots \(\pm i\) come from the quadratic factor \(r^2 + 1 = 0\). Complex roots usually imply an oscillatory component in the solution, which is critical in fields involving wave or vibrational analysis.When you have these roots, the general solution will involve terms that include both sine and cosine functions which correspond to the imaginary unit \(i\). Understanding these roots helps predict the behavior of the system over time, especially how it oscillates around its equilibrium position.

General Solution

The general solution of a differential equation is a combination of solutions corresponding to each root of the characteristic equation. When you have different types of roots like real and complex, you need to include all in the solution.In this problem, the roots were \(r_1 = 1\) and \(r_2, r_3 = \pm i\). Hence, the general solution is:

• \(y(t) = c_1 e^t + c_2 e^{-it} + c_3 e^{it}\)

Here, \(c_1, c_2,\) and \(c_3\) are constants obtained through initial conditions. The terms with \(e^{it}\) and \(e^{-it}\) represent the oscillatory part due to complex roots. This solution structures the broader response of the differential equation, serving as the blueprint before setting specific conditions.

Specific Solution

The specific solution is obtained by applying the initial conditions of the problem to the general solution. It personalizes the general response to meet particular needs of the differential equation scenario. For our problem, we have:

• \(y(0) = 2\)
• \(y^{\prime}(0) = -1\)
• \(y^{\prime\prime}(0) = -2\)

Using these initial conditions, we solve for \(c_1, c_2,\) and \(c_3\), resulting in values \(c_1 = 1\), \(c_2 = 1+i\), and \(c_3 = 1-i\). These constants adjust the general solution to:

• \(y(t) = e^t + (1+i)e^{-it} + (1-i)e^{it}\)

This specific solution provides a full picture, considering both the natural response and forced response conditions set by the problem's initial parameters. This translates into the actual behavior model of the scenario the differential equation depicts.