Question

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

Short answer

Question: Find the general solution of the third-order linear differential equation $$y^{\prime \prime \prime}-y^{\prime} = e^{2t}$$. Answer: The general solution of the given third-order linear differential equation is $$y(t) = C_1 + C_2e^t + C_3e^{-t} + \frac{t}{42}e^{2t}$$.

Step-by-step solution

Find the complementary solution

To find the complementary solution, first, solve the homogeneous equation $$y^{\prime \prime \prime}-y^{\prime} = 0$$. Assuming the solution is of the form \(y = e^{rt}\), we substitute it into the equation and obtain the associated characteristic equation: $$r^3 - r = 0$$ Factoring, we get: $$r(r^2 - 1) = r(r-1)(r+1) = 0$$ Hence, there are three real roots, \(r_1=0\), \(r_2=1\), and \(r_3=-1\). Consequently, the complementary solution is given by: $$y_c(t) = C_1 + C_2e^t + C_3e^{-t}$$

Find a particular solution

For the given nonhomogeneous equation, the right side is of the form $$e^{2t}$$. We'll assume a particular solution of the form $$y_p(t) = Ate^{2t}$$, where \(A\) is a coefficient to be determined. Now, compute the first, second, and third derivatives: $$y_p^{\prime}(t) = A(2e^{2t} + 4te^{2t})$$ $$y_p^{\prime\prime}(t) = A(12e^{2t} + 16te^{2t})$$ $$y_p^{\prime\prime\prime}(t) = A(44e^{2t} + 32te^{2t})$$ Substituting \(y_p(t)\) and its derivatives into the given nonhomogeneous equation, we get: $$A(44e^{2t} + 32te^{2t}) - A(2e^{2t} + 4te^{2t}) = e^{2t}$$ This simplifies to: $$A(42e^{2t} + 28te^{2t}) = e^{2t}$$ Comparing the coefficients of matching powers of \(t\), we find: $$42A = 1$$ Thus, \(A = \frac{1}{42}\). So, the particular solution is: $$y_p(t) = \frac{t}{42}e^{2t}$$

Formulate the general solution

Finally, we combine the complementary and particular solutions to find the general solution: $$y(t) = y_c(t) + y_p(t) = C_1 + C_2e^t + C_3e^{-t} + \frac{t}{42}e^{2t}$$

Key concepts

Complementary Solution

In differential equations, the complementary solution represents the solution to the related homogeneous equation. The homogeneous equation is formed by removing any non-homogeneous terms, in this instance, transforming the original equation to \(y^{\prime \prime \prime}-y^{\prime}=0\). The focus is on finding solutions that satisfy this simpler form.

To find the complementary solution, we assume a potential solution form \(y = e^{rt}\), where \(r\) is a constant. By substituting into the homogeneous equation, we derive a characteristic equation, in this case, \(r^3 - r = 0\). This characteristic equation represents a polynomial in terms of \(r\), which we solve to determine the roots.

• For our equation, factoring results in \(r(r^2 - 1) = 0\), yielding roots \(r_1 = 0\), \(r_2 = 1\), and \(r_3 = -1\).
• The roots indicate the exponential parts of the solution: \(e^{0t}\), \(e^{t}\), and \(e^{-t}\), respectively.

Thus, the complementary solution is a combination of these fundamental solutions: \(y_c(t) = C_1 + C_2e^t + C_3e^{-t}\), where \(C_1, C_2,\) and \(C_3\) are arbitrary constants determined by boundary conditions.

Particular Solution

The particular solution of a differential equation tackles the problem's non-homogeneous part. This is the portion of the solution that ensures the differential equation equals the non-zero function present in the equation.

For our given equation \(y^{\prime \prime \prime}-y^{\prime}=e^{2t}\), the particular solution matches the exponential function on the right side of the equation. Here, we utilize a method called "the method of undetermined coefficients." This involves assuming a solution form that mirrors the non-homogeneous part. We assumed \(y_p(t) = Ate^{2t}\), where \(A\) is a coefficient to be determined.

• We calculate derivatives of \(y_p(t)\) and substitute them back into the differential equation.
• Matching the coefficients of like terms allows us to solve for \(A\), leading to the particular solution \(y_p(t) = \frac{t}{42}e^{2t}\).

This particular solution addresses the specific non-homogeneous nature of the equation, contributing crucially to forming the general solution.

General Solution

The general solution incorporates both the complementary and particular solutions. It represents the full set of solutions that satisfy the original differential equation, taking into account both the homogeneous and non-homogeneous components.

For our differential equation, we bring together the complementary solution \(y_c(t) = C_1 + C_2e^t + C_3e^{-t}\) and the particular solution \(y_p(t) = \frac{t}{42}e^{2t}\).

• The general solution is expressed as a sum: \(y(t) = y_c(t) + y_p(t)\).
• Thus, \(y(t) = C_1 + C_2e^t + C_3e^{-t} + \frac{t}{42}e^{2t}\).

This equation captures all potential solutions to the original problem, incorporating both the arbitrary nature of the constants from the complementary solution and the specificity of the non-homogeneous effect covered by the particular solution. Crafting the general solution is crucial for solving practical problems, where initial or boundary conditions could help determine specific values for \(C_1, C_2,\) and \(C_3\).