Question

For each differential equation, (a) Find the complementary solution. (b) Find a particular solution. (c) Formulate the general solution. $$ y^{\prime \prime \prime}+y^{\prime \prime}=6 e^{-t} $$

Short answer

Question: Determine the general solution of the following differential equation: \(y^{\prime \prime \prime} + y^{\prime \prime} = 6e^{-t}\). Answer: The general solution is given by: \(y(t) = C_1 + C_2e^{-t} - 6te^{-t}\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Find the complementary solution

To find the complementary solution, first, we need to solve the homogeneous differential equation associated with the given differential equation. The homogeneous equation is: $$ y^{\prime \prime \prime} + y^{\prime \prime} = 0 $$ This is a linear differential equation with constant coefficients, so we can solve it by looking for solutions of the form: $$ y(t) = e^{rt} $$ where r is a constant. Differentiating and plugging back into the equation, we have: $$ r^3 e^{rt} + r^2 e^{rt} = 0 $$ This can be factored as follows: $$ r^2 e^{rt}(r + 1) = 0 $$ This equation holds for r = 0 and r = -1. Therefore, the complementary solution has the form: $$ y_c(t) = C_1 + C_2e^{-t} $$ where \(C_1\) and \(C_2\) are constants.

Find a particular solution

For the inhomogeneous differential equation, we want to find a particular solution. Since the right-hand side is a multiple of \(e^{-t}\), we can guess the particular solution of the form: $$ y_p(t) = Ate^{-t} + Be^{-t} $$ Differentiating \(y_p(t)\) gives: $$ y_p^\prime(t) = -Ate^{-t} - Be^{-t} + Ae^{-t} $$ and $$ y_p^{\prime \prime}(t) = Ate^{-t} + 2Ae^{-t} + Be^{-t} $$ Differentiating \(y_p(t)\) once more gives: $$ y_p^{\prime \prime \prime}(t) = -Ate^{-t} - 3Ae^{-t} - Be^{-t} $$ Now, substitute the expressions for \(y^{\prime \prime \prime}\), \(y^{\prime \prime}\), and \(y_p\) into the differential equation: $$ (-Ate^{-t} - 3Ae^{-t} - Be^{-t}) + (Ate^{-t} + 2Ae^{-t} + Be^{-t}) = 6e^{-t} $$ This simplifies to: $$ -Ae^{-t} = 6e^{-t} $$ Solving for the constants A and B, we get: $$ A = -6 $$ Thus, the particular solution is: $$ y_p(t) = -6te^{-t} $$

Formulate the general solution

The general solution is the sum of the complementary solution and the particular solution: $$ y(t) = y_c(t) + y_p(t) = C_1 + C_2e^{-t} - 6te^{-t} $$ This is the general solution for the given differential equation.

Key concepts

Complementary Solution

In solving differential equations, the complementary solution is a key component, especially for linear differential equations with constant coefficients. The complementary solution represents the solution of the associated homogeneous equation, which in this case, removes the non-homogeneous part to solve separately.To find it, we look at the homogeneous version of the equation: \[ y^{'''} + y^{''} = 0 \] For this, we assume a trial solution of the form \( y(t) = e^{rt} \) where \( r \) is a constant. By substituting this into the differential equation, we get: \[ r^3 e^{rt} + r^2 e^{rt} = 0 \] which simplifies to \[ r^2 e^{rt}(r + 1) = 0 \].This yields the characteristic equation \( r^2 (r + 1) = 0 \), giving roots \( r = 0 \) and \( r = -1 \). Hence, the complementary solution is: \[ y_c(t) = C_1 + C_2e^{-t} \] where \( C_1 \) and \( C_2 \) are arbitrary constants. This solution accounts for the natural response of the system without any external forcing term.

Particular Solution

Finding the particular solution involves tackling the inhomogeneous part of a differential equation. This solution will satisfy the entire non-homogeneous equation, given that a complementary solution satisfies the homogeneous equation.For the original differential equation: \[ y^{'''}+y^{''}=6e^{-t} \] we guess a form for the particular solution. Because the non-homogeneous term is \( 6e^{-t} \), we try a solution of similar form: \[ y_p(t) = Ate^{-t} + Be^{-t} \].By differentiating, we find:

• \( y_p^ ext{'}(t) = -Ate^{-t} - Be^{-t} + Ae^{-t} \)
• \( y_p^{''}(t) = Ate^{-t} + 2Ae^{-t} + Be^{-t} \)
• \( y_p^{'''}(t) = -Ate^{-t} - 3Ae^{-t} - Be^{-t} \)

Substituting these into the original equation simplifies to: \[ -Ae^{-t} = 6e^{-t} \]. Solving gives \( A = -6 \), hence the particular solution becomes: \[ y_p(t) = -6te^{-t} \].This particular solution will "cancel out" the inhomogeneous part when substituted back into the original differential equation.

Linear Differential Equation with Constant Coefficients

Linear differential equations with constant coefficients are a fundamental type of differential equation that appear frequently in mathematics, physics, and engineering.These equations have the characteristic feature that each term is either a derivative of the function or the function itself, multiplied by a constant. By examining equations such as the given: \[ y^{'''}+y^{''}=6e^{-t} \] we note this type meets the criteria for constant coefficients because the coefficients of each term (before the sake of the derivative) are consistent or do not change with \( t \).The method to solve them involves complementary and particular solutions:

• Complementary Solution: Solves the homogeneous equation \( y^{'''} + y^{''} = 0 \) by finding roots of the characteristic equation (here, \( r^2(r + 1) = 0 \)) and then building a solution from combinations of these roots' exponential functions.
• Particular Solution: Tackles the equation with the non-homogeneous side using various techniques—sometimes guessing a particular function form based on the non-homogeneous part is needed.

Solving such differential equations provides insights into the natural behaviors of systems as well as their responses to external stimuli.