Question

Solve the initial value problem and graph the solution. $$ y^{\prime \prime \prime}-y^{\prime \prime}-y^{\prime}+y=-e^{-x}(4-8 x), \quad y(0)=2, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=0 $$

Short answer

Question: Find the specific solution for the initial value problem (IVP) given by the third order linear differential equation with constant coefficients and initial conditions: $$ y^{\prime \prime \prime} - y^{\prime \prime} - y^{\prime} + y = -e^{-x}(4-8x), \quad y(0) = 2, \quad y^{\prime}(0) = 0, \quad y^{\prime \prime}(0) = 0 $$ Also, provide the graph of the specific solution. Answer: The specific solution for the given IVP is: $$ y(x) = \frac{15}{2}e^x + \cos x + \sin x -\frac{9}{2}e^{-x}(x^2-3x+\frac{17}{4}) $$ To graph the specific solution, you can use appropriate graphing software (for example, Desmos, GeoGebra, or Mathematica) to plot the function: $$ y(x) = \frac{15}{2}e^x + \cos x + \sin x -\frac{9}{2}e^{-x}(x^2-3x+\frac{17}{4}) $$ The graph should have initial values at (0,2), and the curve should have a relative balance between exponential growth and oscillatory behavior.

Step-by-step solution

Find the complementary function

To find the complementary function, we need to solve the associated homogeneous equation: $$ y^{\prime \prime \prime} - y^{\prime \prime} - y^{\prime} + y = 0 $$ For homogeneous equations with constant coefficients, we assume a solution of the form \(y=e^{r x}\). Substituting this into the equation and dividing by \(e^{rx}\), we get the characteristic equation: $$ r^3 - r^2 - r + 1 = 0 $$ We can rewrite this as \((r-1)(r^2+1)=0\). This gives us three roots: \(r=1\), \(r=i\) and \(r=-i\). So, the complementary function is: $$ y_c(x) = C_1e^x + C_2 \cos x + C_3 \sin x $$ Where \(C_1\), \(C_2\), and \(C_3\) are constants.

Find the particular solution

Since the non-homogeneous part is of the form \(-e^{-x}(4-8x)\), we assume a particular solution of the form: $$ y_p(x) = e^{-x}(Ax^2+Bx+C) $$ Now, we'll find the first, second and third derivatives of \(y_p(x)\): $$ y_p^{\prime}(x) = e^{-x}(-Ax^2+(2A-B)x+(A-B+C)) $$ $$ y_p^{\prime \prime}(x) = e^{-x}(Ax^2-(4A-B)x+(6A-2B+C)) $$ $$ y_p^{\prime \prime \prime}(x) = e^{-x}(-Ax^2+(6A-2B)x-(12A+2B+C)) $$ Substitute \(y_p(x)\), \(y_p^{\prime}(x)\), \(y_p^{\prime \prime}(x)\), and \(y_p^{\prime \prime \prime}(x)\) into the non-homogeneous equation and equate coefficients: $$ (-A+1)=-4, \quad (6A-2B)=(4B+8), \quad (-12A+2B+C)=-8 $$ Solve these equations to get \(A=-\frac{9}{2}\), \(B=3\), and \(C=\frac{17}{2}\). So, the particular solution is: $$ y_p(x) = -\frac{9}{2}e^{-x}(x^2-3x+\frac{17}{4}) $$

The general solution

The general solution is the sum of the complementary function and the particular solution: $$ y(x) = y_c(x) + y_p(x) = C_1e^x + C_2 \cos x + C_3 \sin x -\frac{9}{2}e^{-x}(x^2-3x+\frac{17}{4}) $$

Use the initial conditions

Now we need to find the specific solution using the initial conditions: $$ y(0) = 2 \Rightarrow C_1 + C_2 = \frac{17}{2} $$ $$ y^{\prime}(0) = 0 \Rightarrow C_1 - C_2 + 3C_3 = 0 $$ $$ y^{\prime \prime}(0) = 0 \Rightarrow C_1 - C_2 - 2C_3 = 9 $$ Solving these equations gives us \(C_1 = \frac{15}{2}\), \(C_2 = 1\), and \(C_3=1\). The specific solution is: $$ y(x) = \frac{15}{2}e^x + \cos x + \sin x -\frac{9}{2}e^{-x}(x^2-3x+\frac{17}{4}) $$

Graph the specific solution

To graph the specific solution, you can use appropriate graphing software (for example, Desmos, GeoGebra, or Mathematica) to plot the function: $$ y(x) = \frac{15}{2}e^x + \cos x + \sin x -\frac{9}{2}e^{-x}(x^2-3x+\frac{17}{4}) $$ You should see a graph with initial values at \((0,2)\), and the curve should have a relative balance between exponential growth and oscillatory behavior.

Key concepts

Initial Value Problem

An initial value problem in the realm of differential equations is a type of problem where the solution to a differential equation needs to pass through a specific point or set of points. These points are known as initial conditions. In simpler terms, an initial value problem requires the solution not only to satisfy the differential equation but also to start from a certain known state.

For instance, in the given example, the differential equation is third-order, meaning it involves the third derivative. The initial conditions provided are:

• The function value at zero, \(y(0) = 2\)
• The first derivative at zero, \(y'(0) = 0\)
• The second derivative at zero, \(y''(0) = 0\)

These initial conditions help specifically define the unique solution for the problem among the many possible solutions to the differential equation.

Complementary Function

When dealing with linear differential equations, the complementary function plays a crucial role. This function is the solution to the associated homogeneous equation. Homogeneous equations are equations where the solution equals zero.

For the problem at hand, the complementary function is derived from the homogeneous equation:\[ y''' - y'' - y' + y = 0\]We assume a solution structure involving exponential and trigonometric functions: \(y = e^{rx}\). By substituting and solving for \(r\), we identify the characteristic equation, revealing the roots: \(1\), \(i\), and \(-i\).

These roots guide the construction of the complementary function, resulting in:\[y_c(x) = C_1e^x + C_2 \, \cos x + C_3 \, \sin x\]Here, \(C_1\), \(C_2\), and \(C_3\) are constants to be determined using the initial conditions.

Particular Solution

The particular solution is a vital component when solving non-homogeneous differential equations. It accounts for the non-zero part of the equation. The method of undetermined coefficients is frequently used to find it. In this approach, a guess is tailored based on the function on the right side of the differential equation.

In our example, where the non-homogeneous part involves \(-e^{-x}(4-8x)\), we hypothesize a particular solution: \( y_p(x) = e^{-x}(Ax^2 + Bx + C)\).Through differentiation and substitution back into the non-homogeneous differential equation, we solve for the coefficients \(A\), \(B\), and \(C\). This gives the specific form of our particular solution:\[y_p(x) = -\frac{9}{2}e^{-x}(x^2 - 3x + \frac{17}{4})\]The particular solution complements the complementary function in modeling the complete problem.

Non-Homogeneous Equation

Non-homogeneous equations include an extra term not present in homogeneous equations, typically represented as a function of \(x\) that is not zero. This term distinguishes non-homogeneous equations and requires the solution of both the complementary function and the particular solution.

In this exercise, the non-homogeneous equation is expressed as:\[y''' - y'' - y' + y = -e^{-x}(4 - 8x)\]To tackle it, we solve the corresponding homogeneous equation to find the complementary function and use methods like undetermined coefficients for the particular solution.

Ultimately, the solution to the non-homogeneous differential equation is the sum of these parts, resulting in:\[y(x) = y_c(x) + y_p(x)\]This cumulative solution honors both the structure of the equation and the specific initial conditions provided.