Question

Determine the general solution of the given differential equation. \(y^{\prime \prime \prime}-y^{\prime}=2 \sin t\)

Short answer

Question: Find the general solution of the third-order linear inhomogeneous differential equation \(y^{\prime\prime\prime}-y^{\prime}=2\sin t\). Answer: The general solution of the given differential equation is \(y(t)=C_1+C_2e^t+C_3e^{-t}-\sin t+\cos t\).

Step-by-step solution

Identify complementary function (homogeneous equation)

Find the complementary function of the homogeneous equation: \(y^{\prime \prime \prime}-y^{\prime}=0\) The characteristic equation is: \(r^3-r=0\) Factor out \(r\). We get: \(r(r^2-1)=0\) This equation has three real roots: \(r=0,1,-1\) Therefore, the complementary function is: \(y_c(t)=C_1+C_2e^t+C_3e^{-t}\), where \(C_1\), \(C_2\), and \(C_3\) are constants.

Find a particular solution (inhomogeneous equation)

Now we need to find a particular solution to the inhomogeneous equation: \(y^{\prime \prime \prime}-y^{\prime}=2 \sin t\) Guess a solution of the form: \(y_p(t)=A\sin t+B\cos t\) The first derivative is: \(y_p^{\prime}(t)=A\cos t-B\sin t\) The second derivative is: \(y_p^{\prime\prime}(t)=-A\sin t-B\cos t\) The third derivative is: \(y_p^{\prime\prime\prime}(t)=-A\cos t+B\sin t\) Now substitute \(y_p\), \(y_p^{\prime}\), and \(y_p^{\prime\prime\prime}\) into the inhomogeneous equation and equate the coefficients of \(\sin t\) and \(\cos t\). For \(\sin t\): \((-A\cos t+B\sin t)(\sin t)-(A\cos t-B\sin t)(\cos t)=2\sin t\) \(-A\sin^2 t + B\sin t\cos t -A\cos^2 t + B\sin t\cos t = 2\sin t\) Now equate coefficients of \(\sin t\) and \(\cos t\): For \(\sin t\): \(-A+B=2\) For \(\cos t\): \(-A+B=0\) Solve the system of equations, we get: \(A=-1\) \(B=1\) So, the particular solution is: \(y_p(t)=-\sin t+\cos t\)

Find the general solution

Combine the complementary function and the particular solution to find the general solution of the given differential equation: \(y(t)=y_c(t)+y_p(t)\) Hence, the general solution is: \(y(t)=C_1+C_2e^t+C_3e^{-t}-\sin t+\cos t\) This is the general solution of the given differential equation.

Key concepts

Third-Order Differential Equation

A third-order differential equation is a type of differential equation in which the highest derivative is the third derivative. These types of equations can model various physical phenomena, such as vibrations and wave propagation. In the equation \(y''' - y' = 2\sin t\), we are dealing with a third-order equation because the term \(y'''\) signifies the third derivative of the function \(y\) with respect to \(t\). Here are some characteristics:

• The degree of the equation is dictated by the highest derivative involved, which in this case is three, making it a third-order equation.
• Solving such equations often involves finding both the complementary function and the particular solution.
• It typically requires understanding the underlying behavior of the system modeled by the equation.

Solving this equation involves using techniques aimed at breaking it down into manageable parts, which makes understanding each component essential for finding a valid solution.

Characteristic Equation

The characteristic equation is a crucial step in solving linear differential equations, especially with constant coefficients. In our example for the homogeneous equation \(y''' - y' = 0\), creating a characteristic equation helps to determine the complementary function. The characteristic equation is formed by substituting \(r\) for each derivative, ending up with \(r^3 - r = 0\). Here’s a closer look at the process:

• The characteristic equation itself \(r^3 - r = 0\) can be factored to \(r(r^2 - 1) = 0\).
• This results in finding roots of the equation: \(r = 0\), \(r = 1\), and \(r = -1\).
• These roots are key to constructing the complementary function, which represents the general solution to the homogeneous part of the differential equation.

The characteristic equation is an essential tool for visualizing the behavior of solutions and understanding their properties based on the roots obtained.

Complementary Function

The complementary function (CF) arises from solving the homogeneous part of a differential equation. It reflects the natural behavior of the system without external forcing components. For the homogeneous equation \(y''' - y' = 0\), after determining the roots \(r = 0, 1, -1\), we can build the complementary function as \(y_c(t) = C_1 + C_2e^t + C_3e^{-t}\).

• The CF is assembled using the roots from the characteristic equation, each contributing a component (e.g., \(e^t\) for real distinct roots).
• For a root \(r = 0\), it contributes a constant \(C_1\) to the CF.
• Other roots lead to exponential terms, forming products with constants \(C_2\) and \(C_3\).

Each part of the CF represents a solution that satisfies the differential equation without any nonhomogeneous terms, forming a critical part of the general solution.

Particular Solution

A particular solution (PS) is needed to account for the nonhomogeneous part of a differential equation, which contains the external driving force or function, such as \(2\sin t\) in our case. The goal is to find a specific solution that fits the entire differential equation. To achieve this, we use the method of undetermined coefficients by guessing a form: \(y_p(t) = A\sin t + B\cos t\). This is because the nonhomogeneous term is sinusoidal.

• The chosen form emulates the nonhomogeneous part, which is \(2\sin t\), targeting specific terms.
• You then derive and substitute this assumed solution back into the equation.
• By matching coefficients from both sides for \(\sin t\) and \(\cos t\), you find specific values for coefficients \(A\) and \(B\), achieving \(A = -1\) and \(B = 1\).

Combining this particular solution with the complementary function gives the broad picture of how the differential equation solves in terms of both inherent system behavior and external forces.