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. $$ y^{\prime \prime}-5 y^{\prime}+6 y=2 e^{t} $$

Short answer

Question: Find the particular solution to the given second-order differential equation using variation of parameters and undetermined coefficients methods: $$y^{\prime \prime}-5 y^{\prime}+6 y= 2e^{t}$$ Answer: The particular solution is: $$y_p = \frac{5}{3}$$

Step-by-step solution

Find the characteristic equation and roots

The characteristic equation of this homogeneous equation is: $$r^2-5r+6=0$$ This quadratic equation can be factored as follows: $$(r-2)(r-3)=0$$ So, the roots are r = 2, 3.

Write the complementary solution

Our complementary solution is a linear combination of our root solutions: $$y_h = C_1 e^{2t} + C_2 e^{3t}$$ Now, let's move on to finding the particular solution using the variation of parameters method.

Find the Wronskian

The Wronskian, W, is used to find the coefficients of our particular solution using the variation of parameters formula: $$W=\begin{vmatrix} e^{2t} & e^{3t}\\ e^{2t} (2) & e^{3t} (3)\end{vmatrix}= 3e^{5t}-2e^{5t}=e^{5t}$$

Calculate u_1 and u_2

Using the variation of parameters formulas, we'll integrate to find u_1 and u_2: $$u_1=\int \frac{- ye^{3t}dt}{W} = \int -2e^{-2t} dt = e^{-2t} + C_3$$ $$u_2=\int \frac{ ye^{2t}dt}{W} =\int 2e^{-3t} dt = \frac{2}{3}e^{-3t} + C_4$$ Since we are looking for the particular solution, the constants of integration can be set to 0: $$u_1 = e^{-2t}$$ $$u_2 = \frac{2}{3}e^{-3t}$$

Form the particular solution

Now, we can write our particular solution y_p using u_1 and u_2: $$y_p = u_1 y_1 + u_2 y_2 = e^{-2t} e^{2t} + \frac{2}{3} e^{-3t} e^{3t} = 1 + \frac{2}{3}$$ $$y_p = \frac{5}{3}$$ Now, let's verify this solution using the method of undetermined coefficients.

Choose the form of particular solution using undetermined coefficients

The forcing function \(2e^t\) suggests that the particular solution must have the form: $$y_p = Ae^t$$

Differentiate and substitute into the inhomogeneous equation

We need to differentiate \(y_p\) and substitute it into the inhomogeneous equation: $$y_p^{\prime} = At e^t$$ $$y_p^{\prime \prime} = At e^t + Ae^t$$ Substitute the terms into the inhomogeneous equation: $$\left( At e^t + Ae^t \right) - 5 \left( At e^t \right) + 6 \left( Ae^{t} \right) = 2 e^{t}$$

Solve for A

Now, we equate coefficients to solve for A: $$\left( 1 + A -5A +6A \right) e^t = 2 e^t$$ $$\Rightarrow A = \frac{5}{3}$$ So our particular solution using undetermined coefficients is: $$y_p = \frac{5}{3} e^{t}$$ Since we've arrived at the same answer for our particular solution using both methods, we have successfully checked our answer. Our final solution for the given differential equation is: $$y = y_h + y_p = C_1 e^{2t} + C_2 e^{3t} + \frac{5}{3}$$