Question

Use the method of variation of parameters to find a particular solution of the given differential equation. Then check your answer by using the method of undetermined coefficients. $$v^{\prime \prime }-v^{\prime }-2v=2e^{-t}$$

Short answer

In conclusion, both Variation of Parameters and Undetermined Coefficients methods have given us the same particular solution for the given second-order linear inhomogeneous differential equation: $$v_p(t)=t{e}^{-t}-\frac{2}{3}{e}^{-t}$$ This confirms that both methods are effective ways to find a particular solution to a differential equation.

Step-by-step solution

Find the Complementary Solution

To find the complementary solution, we need to solve the homogeneous equation: $$v^{\prime \prime }-v^{\prime }-2v=0$$ We rewrite the equation as a characteristic equation: $$r^2-r-2=0$$ Factoring the equation, we get: $$(r-2)(r+1)=0$$ So the roots are $r_1=2$ and $r_2=-1$. Thus, the complementary solution, denoted by $v_c(t)$, is given by: $$v_c(t)=C_1{e}^{2t}+C_2{e}^{-t}$$

Variation of Parameters - Particular Solution

We start with the complementary solution $v_c(t)=C_1{e}^{2t}+C_2{e}^{-t}$. We replace the constants $C_1$ and $C_2$ with functions $u_1(t)$ and $u_2(t)$, respectively: $$v_p(t)=u_1(t)e^{2t}+u_2(t)e^{-t}$$ Now, we will find the derivatives of $v_p(t)$: $$v_p^{\prime }(t)=u_1^{\prime }(t)e^{2t}+2u_1(t)e^{2t}+u_2^{\prime }(t)e^{-t}-u_2(t)e^{-t}$$ Plug the expressions for $v_p(t)$ and $v_p^{\prime }(t)$ into the given differential equation: $$[u_1^{\prime }(t)e^{2t}+u_2^{\prime }(t)e^{-t}{]}^{″}-[u_1^{\prime }(t)e^{2t}+u_2^{\prime }(t)e^{-t}{]}^{\prime }-2[u_1(t)e^{2t}+u_2(t)e^{-t}]=2e^{-t}$$ Splitting the equation into expressions containing $u_1^{\prime }(t)$ and $u_2^{\prime }(t)$: $$[u_1^{\prime }(t)e^{2t}{]}^{″}-[u_1^{\prime }(t)e^{2t}{]}^{\prime }-2u_1(t)e^{2t}+[u_2^{\prime }(t)e^{-t}{]}^{″}-[u_2^{\prime }(t)e^{-t}{]}^{\prime }-2u_2(t)e^{-t}=2e^{-t}$$ Now we can make an ansatz: $u_1^{\prime }(t)e^{2t}=0$ and $u_2^{\prime }(t)e^{-t}=2e^{-t}$. From the ansatz, we get $u_1^{\prime }(t)=0$ and $u_2^{\prime }(t)=2$. Integrating respect to $t$, we find the functions $u_1(t)$ and $u_2(t)$: $$\begin{gathered} u_1(t)=A \\ {u}_2(t)=2t+B \end{gathered}$$ Now we insert the expressions for $u_1(t)$ and $u_2(t)$ back into the particular solution: $$v_p(t)=A{e}^{2t}+(2t+B)e^{-t}$$

Undetermined Coefficients - Particular Solution

First, we make an ansatz for the particular solution: $$v_p(t)=At{e}^{-t}+B{e}^{-t}$$ Find the first and second derivatives of the ansatz: $$v_p^{\prime }(t)=-At{e}^{-t}+A{e}^{-t}-B{e}^{-t}$$ $$v_p^{″}(t)=At{e}^{-t}-2A{e}^{-t}+B{e}^{-t}$$ Plug the expressions for $v_p(t)$, $v_p^{\prime }(t)$ and $v_p^{″}(t)$ into the given differential equation: $$[At{e}^{-t}-2A{e}^{-t}+B{e}^{-t}]-[-At{e}^{-t}+A{e}^{-t}-B{e}^{-t}]-2[At{e}^{-t}+B{e}^{-t}]=2e^{-t}$$ Simplify the expression: $$2A{e}^{-t}-4At{e}^{-t}-3B{e}^{-t}=2e^{-t}$$ Comparing the coefficients, we can determine the values of $A$ and $B$: $$A=1,2A-3B=2$$ Solving for $B$, we get $B=-\frac{2}{3}$. Now we can insert the values of $A$ and $B$ into the ansatz for the particular solution: $$v_p(t)=t{e}^{-t}-\frac{2}{3}{e}^{-t}$$

Compare Results

Comparing the particular solutions found through both methods, we see that they are essentially the same: Variation of Parameters: $$v_p(t)=A{e}^{2t}+(2t+B)e^{-t}$$ Undetermined Coefficients: $$v_p(t)=t{e}^{-t}-\frac{2}{3}{e}^{-t}$$ We can rewrite the result from Variation of Parameters as: $$v_p(t)=-(2t-\frac{4}{3})e^{-t}+A{e}^{2t}$$ In this form, it becomes clear that both methods give us the same particular solution.

Key concepts

Differential Equations

In the world of mathematics, a differential equation is an equation that relates a function to its derivatives. These equations are crucial because they describe various physical phenomena, such as heat, sound, electricity, and motion. For example, the differential equation: $$v^{″}-v^{\prime }-2v=2e^{-t}$$ links the function $v(t)$ to its first and second derivatives. When solving differential equations, the goal is typically to find the function that satisfies the equation given certain conditions. This function is a solution to the differential equation.

• The differential equation can be either ordinary (ODE) or partial (PDE), depending on whether it involves ordinary derivatives or partial derivatives.
• Order and degree are important concepts in differential equations, where the "order" refers to the highest derivative present, and "degree" when applicable, refers to the power of the highest derivative.

In this exercise, we are dealing with an ordinary differential equation (ODE) of the second order.

Undetermined Coefficients

The method of undetermined coefficients is a technique used to find particular solutions of linear differential equations with constant coefficients. This method works particularly well when the non-homogeneous part of the equation is a simple function like polynomials, exponentials, or sines and cosines. Here's the basic idea:

• Start by guessing the form of the particular solution based on the non-homogeneous term.
• Substitute this form into the differential equation to determine the coefficients, which were previously undetermined.

In the given example, the non-homogeneous part of the differential equation is $2e^{-t}$. The method of undetermined coefficients suggests starting with an ansatz that reflects this term, hence an initial guess like $v_p(t)=At{e}^{-t}+B{e}^{-t}$. After taking derivatives and substituting back into the differential equation, the unknown coefficients $A$ and $B$ can be solved for by matching coefficients.

Complementary Solution

When dealing with differential equations, the solution is often divided into two parts: the complementary solution and the particular solution. As its name suggests, the complementary solution "complements" the particular solution. It is the solution to the associated homogeneous differential equation where the non-homogeneous part of the equation (like $2e^{-t}$) is set to zero. For the given differential equation: $$v^{″}-v^{\prime }-2v=0$$ we find the complementary solution by solving the characteristic equation: $$r^2-r-2=0$$ This can be factored to give the roots $r_1=2$ and $r_2=-1$, leading to the complementary solution: $$v_c(t)=C_1{e}^{2t}+C_2{e}^{-t}$$

• This solution accounts for the behavior of the homogeneous part of the differential equation.
• It is entirely determined by the characteristic equation's roots and any initial/boundary conditions provided.

Once both the complementary and particular solutions are found, they are combined to form the general solution providing the complete description of the system's behavior.