Question

For each differential equation, (a) Find the complementary solution. (b) Find a particular solution. (c) Formulate the general solution. $$ y^{\prime \prime \prime}-y^{\prime \prime}=4 e^{-2 t} $$

Short answer

Answer: The general solution to the given third-order linear ordinary differential equation is \(y(t) = C_1 + C_2 t + C_3 e^t - \frac{1}{3} e^{-2t}\) where \(C_1\), \(C_2\), and \(C_3\) are constants.

Step-by-step solution

Find the complementary solution

For the complementary solution, we will deal with the homogeneous equation by removing the right-hand side and considering the equation: $$ y^{\prime \prime \prime}-y^{\prime \prime}=0 $$ To find the complementary solution, let's assume a solution of the form \(y(t) = e^{rt}\). Plugging this into the homogeneous equation, we get: $$ r^3 e^{rt} - r^2 e^{rt} = 0 $$ Factor out \(e^{rt}\) and \(r^2\), we have: $$ r^2 e^{rt} (r-1) = 0 $$ The roots are \(r = 0\) (a double root) and \(r = 1\). The complementary solution is then: $$ y_c(t) = C_1 + C_2 t + C_3 e^t $$ where \(C_1\), \(C_2\), and \(C_3\) are constants.

Find a particular solution

For the particular solution, we look for a solution of the form: $$ y_p(t) = A e^{-2t} $$ where \(A\) is a constant to be determined. We need to find the first, second, and third derivatives of \(y_p\): $$ y_p^{\prime}(t) = -2A e^{-2t} $$ $$ y_p^{\prime \prime}(t) = 4A e^{-2t} $$ $$ y_p^{\prime \prime \prime}(t) = -8A e^{-2t} $$ Now, substitute these derivatives back into the original equation: $$ (-8A e^{-2t}) - (4A e^{-2t}) = 4 e^{-2t} $$ Simplify and compare coefficients: $$ (-12A)e^{-2t} = 4e^{-2t} $$ $$ -12A = 4 $$ Solve for \(A\): $$ A = -\frac{1}{3} $$ So the particular solution is: $$ y_p(t) = -\frac{1}{3} e^{-2t} $$

Formulate the general solution

As the general solution will be the linear combination of the complementary and particular solutions, we have: $$ y(t) = y_c(t) + y_p(t) = C_1 + C_2 t + C_3 e^t - \frac{1}{3} e^{-2t} $$ This is the general solution to the given third-order linear ordinary differential equation.

Key concepts

Complementary Solution

When dealing with differential equations, one important task is to find the complementary solution, often denoted as \(y_c(t)\). The complementary solution solves the homogeneous version of the differential equation. This means we ignore any terms on the right-hand side of the equation. By focusing only on zeroing inhomogeneities, we simplify to the essence of the equation's intrinsic behavior.

In our example, by setting up the associated homogeneous equation, \(y^{\prime \prime \prime}-y^{\prime \prime}=0\), we use a technique that assumes a solution of the form \(y(t) = e^{rt}\). Plugging this assumed solution into the homogeneous equation allows us to find the roots of the characteristic equation. Factoring gives us the roots, which in this case are \(r = 0\) (a double root) and \(r = 1\).

This results in the complementary solution:\[y_c(t) = C_1 + C_2 t + C_3 e^t\]
Here, \(C_1\), \(C_2\), and \(C_3\) are constants determined by initial conditions or boundary values. The double root \(r = 0\) manifests in the solution as both a constant term and a linear term, while the root \(r = 1\) corresponds to an exponential function \(e^t\). This part of the solution captures the natural, unforced behavior of the system.

Particular Solution

Next, we seek a particular solution \(y_p(t)\) that addresses the non-homogeneous part of the differential equation. This solution is tailored to satisfy the entire given equation, including the non-zero right-hand side term.

For our equation, we propose a particular solution of the form \(y_p(t) = A e^{-2t}\). This candidate function's form reflects the right-hand side of the original equation, \(4 e^{-2 t}\). By choosing a function with a similar expression, we align our solution strategy with the equation's forcing terms.

We find the derivatives of \(y_p(t)\) and substitute them back into the original differential equation:\[y_p^{\prime}(t) = -2A e^{-2t}\]\[y_p^{\prime \prime}(t) = 4A e^{-2t}\]\[y_p^{\prime \prime \prime}(t) = -8A e^{-2t}\]
Substituting these into the differential equation, we solve for \(A\) by equating coefficients, leading to \(A = -\frac{1}{3}\). Thus, the particular solution becomes:\[y_p(t) = -\frac{1}{3} e^{-2t}\]
This part of the solution reflects how the system responds directly to the external forces modeled by the equation.

General Solution

The general solution of a differential equation integrates both the homogeneous and particular solutions. It encompasses both the natural response of the system and its response to external forces.

The general solution is expressed as a sum of the complementary and particular solutions:\[y(t) = y_c(t) + y_p(t)\]
For our equation, it becomes:\[y(t) = C_1 + C_2 t + C_3 e^t - \frac{1}{3} e^{-2t}\]
The complementary solution \(C_1 + C_2 t + C_3 e^t\) accounts for the inherent behavior of the system, while \(-\frac{1}{3} e^{-2t}\) is tailored to the specific forcing terms in the equation.

This general solution provides a complete picture of the system's dynamics, allowing us to predict its behavior given any initial conditions or boundary values. The constants \(C_1\), \(C_2\), and \(C_3\) remain adjustable factors that will be set to meet specific situational requirements.