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. $$ v^{\prime \prime}-v^{\prime}-2 v=2 e^{-t} $$

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: $$v_p(t) = te^{-t} -\frac{2}{3}e^{-t}$$ 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 \(v_p(t)\): $$ v_p'(t) = u_1'(t) e^{2t} + 2u_1(t) e^{2t} + u_2'(t) e^{-t} - u_2(t) e^{-t} $$ Plug the expressions for \(v_p(t)\) and \(v_p'(t)\) into the given differential equation: $$ [u_1'(t)e^{2t} + u_2'(t)e^{-t}]'' - [u_1'(t)e^{2t} + u_2'(t)e^{-t}]' - 2[u_1(t)e^{2t}+u_2(t)e^{-t}] = 2 e^{-t} $$ Splitting the equation into expressions containing \(u_1'(t)\) and \(u_2'(t)\): $$ [u_1'(t)e^{2t}]'' - [u_1'(t)e^{2t}]' - 2u_1(t)e^{2t} + [u_2'(t)e^{-t}]'' - [u_2'(t)e^{-t}]' - 2u_2(t)e^{-t} = 2 e^{-t} $$ Now we can make an ansatz: \(u_1'(t)e^{2t} = 0\) and \(u_2'(t)e^{-t} = 2e^{-t}\). From the ansatz, we get \(u_1'(t) = 0\) and \(u_2'(t) = 2\). Integrating respect to \(t\), we find the functions \(u_1(t)\) and \(u_2(t)\): $$ u_1(t) = A\\ u_2(t) = 2t + B $$ 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: $$v_p(t) = Ate^{-t} + Be^{-t}$$ Find the first and second derivatives of the ansatz: $$v_p'(t) = -Ate^{-t} + Ae^{-t} - Be^{-t}$$ $$v_p''(t) = Ate^{-t} - 2Ae^{-t} + Be^{-t}$$ Plug the expressions for \(v_p(t)\), \(v_p'(t)\) and \(v_p''(t)\) into the given differential equation: $$ [Ate^{-t} - 2Ae^{-t} + Be^{-t}] - [-Ate^{-t} + Ae^{-t} - Be^{-t}] - 2[Ate^{-t} + Be^{-t}] = 2e^{-t} $$ Simplify the expression: $$2Ae^{-t} - 4Ate^{-t}-3Be^{-t}=2e^{-t}$$ 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: $$ v_p(t) = te^{-t} -\frac{2}{3}e^{-t} $$

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: $$ v_p(t) = te^{-t} -\frac{2}{3}e^{-t} $$ 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.