Question

Find the solution of the given initial value problem. Then plot a graph of the solution. $$ y^{\prime \prime \prime}-y^{\prime \prime}+y^{\prime}-y=\sec t, \quad y(0)=2, \quad y^{\prime}(0)=-1, \quad y^{\prime \prime}(0)=1 $$

Short answer

Answer: The exact solution of the given initial value problem is $$y(t)=-\frac{5}{8}e^t+\frac{17}{8}\cos t-\frac{9}{8}\sin t+\frac{1}{2}\cos t + \frac{1}{2}\sin t.$$

Step-by-step solution

Identify the problem type

We are given a third-order linear differential equation$$y^{\prime \prime \prime}-y^{\prime \prime}+y^{\prime}-y=\sec t.$$The corresponding homogeneous equation is$$y^{\prime \prime \prime}-y^{\prime \prime}+y^{\prime}-y=0.$$We will first solve the homogeneous equation and then find a particular solution for the non-homogeneous equation. After that, we will apply the initial conditions to find the exact solution.

Solve the homogeneous equation

The characteristic equation for the homogeneous third-order linear differential equation is$$ r^3-r^2+r-1=0. $$To find the roots of the characteristic equation, we can factor it as follows:$$(r-1)(r^2+1)=0.$$This gives us three distinct roots:$$ r_1 = 1, \quad r_2 = i, \quad r_3 = -i. $$Hence, the general solution for the homogeneous equation is$$ y_h(t)=C_1e^t+C_2\cos t+C_3\sin t, $$where \(C_1\), \(C_2\), and \(C_3\) are constants.

Find a particular solution for the non-homogeneous equation

Since the non-homogeneous equation has the form$$y^{\prime \prime \prime}-y^{\prime \prime}+y^{\prime}-y=\sec t,$$we will use the method of undetermined coefficients to find the particular solution. Assume a particular solution of the form$$ y_p(t)=A\cos t + B\sin t. $$Now, we need to compute the first, second, and third derivatives of \(y_p(t)\):$$ y_p^{\prime}(t)= -A\sin t + B\cos t, \\ y_p^{\prime \prime}(t)= -A\cos t - B\sin t, \\ y_p^{\prime \prime \prime}(t)= A\sin t - B\cos t. $$Substitute these derivatives into the non-homogeneous differential equation:$$ (A\sin t - B\cos t)-(-A\cos t - B\sin t)+(-A\sin t + B\cos t)- (A\cos t + B\sin t) = \sec t. $$Simplifying, we obtain$$ (2B-2A)\cos t +(2A-2B)\sin t = \sec t. $$This implies that$$ 2B-2A=1, \quad 2A-2B=0. $$Solving this system of equations, we find \(A=\dfrac{1}{2}\) and \(B=\dfrac{1}{2}\). Thus, the particular solution is$$ y_p(t)=\frac{1}{2}\cos t + \frac{1}{2}\sin t. $$

Apply the initial conditions

Combine the general solution of the homogeneous equation and the particular solution of the non-homogeneous equation to form the general solution of the given initial value problem:$$ y(t)=y_h(t)+y_p(t)=C_1e^t+C_2\cos t+C_3\sin t+\frac{1}{2}\cos t + \frac{1}{2}\sin t. $$Now, apply the initial conditions \(y(0)=2\), \(y^{\prime}(0)=-1\), and \(y^{\prime \prime}(0)=1\). For \(y(0)=2\):$$ 2=C_1+C_2+\frac{1}{2}. $$For \(y^{\prime}(0)=-1\):$$ -1=C_1+C_3-\frac{1}{2}. $$For \(y^{\prime \prime}(0)=1\):$$ 1=-C_1-C_2+C_3+\frac{1}{2}. $$Solving this system of equations, we find \(C_1=-\dfrac{5}{8}\), \(C_2=\dfrac{17}{8}\), and \(C_3=-\dfrac{9}{8}\). Thus, the exact solution is$$ y(t)=-\frac{5}{8}e^t+\frac{17}{8}\cos t-\frac{9}{8}\sin t+\frac{1}{2}\cos t + \frac{1}{2}\sin t. $$

Plot the graph of the solution

Now that we have found the solution, we can plot the graph of the function \[ y(t)=-\frac{5}{8}e^t+\frac{17}{8}\cos t-\frac{9}{8}\sin t+\frac{1}{2}\cos t + \frac{1}{2}\sin t\] using a graphing tool like Desmos or Geogebra. This graph will represent the behavior of the function and its derivatives over time for the given initial value problem.

Key concepts

Linear Differential Equation

A linear differential equation is an important concept in calculus and differential equations, concerning equations involving derivatives that hold certain linearity properties. In simpler terms, each term in these equations is either a derivative of the function multiplied by a coefficient or simply a function of the independent variable.
The specific linear differential equation provided is third-order, involving the third derivative of function, represented by: - \( y''' - y'' + y' - y = \sec t \).
These equations are called 'linear' because of the absence of powers or products of derivatives. Every order derivative is dealt with separately and linearly, making them more manageable for solutions. Understanding their structure is key to solving them.

Characteristic Equation

The characteristic equation arises when solving homogeneous linear differential equations, providing a route to identify the type of solution the differential equation will have.
For the given problem, the characteristic equation is derived from: - \( r^3 - r^2 + r - 1 = 0 \).This equation is essential for finding the roots which lead to the solution of the homogeneous equation.

• Factor the polynomial: \((r-1)(r^2+1) = 0\).
• Identify roots: \(r_1 = 1, \quad r_2 = i, \quad r_3 = -i \).

These roots are then used to form the general solution of the homogeneous equation, leading towards a complete solution.

Homogeneous Equation

The homogeneous equation is a pivotal part of solving linear differential equations. It represents the portion where the differential equation equals zero, depicted here as: - \( y''' - y'' + y' - y = 0 \).
The general solution derived from the homogeneous equation is constructed using the roots of the characteristic equation. In this scenario:

• Exponential solutions from real roots: \(C_1 e^t\).
• Trigonometric solutions from complex roots: \(C_2 \cos t + C_3 \sin t\).

The overall general solution is then: - \( y_h(t) = C_1 e^t + C_2 \cos t + C_3 \sin t \),where \(C_1, C_2, C_3\) are constants determined later using initial conditions.

Particular Solution

The method of finding a particular solution for a non-homogeneous equation is crucial, as it deals with the other side of the equation, \( \sec t \), in the given example: - \( y''' - y'' + y' - y = \sec t \).
We utilize the method of undetermined coefficients to address this. By assuming a particular solution form like: - \( y_p(t) = A \cos t + B \sin t \),
we substitute into the differential equation and balance coefficients of like terms to solve for \(A\) and \(B\). Solving these gives us \(A = 1/2\) and \(B = 1/2\), resulting in: - \( y_p(t) = \frac{1}{2} \cos t + \frac{1}{2} \sin t \).
Combining the particular solution with the general solution of the homogeneous equation yields the complete solution, tailored to fit the initial conditions provided.