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: $$y = y_h + y_p = C_1 e^{2t} + C_2 e^{3t} + \frac{5}{3}$$

Key concepts

Characteristic Equation

The characteristic equation is a crucial part of solving linear differential equations, especially those with constant coefficients. In this context, it helps find the roots that are used to form the complementary solution. For our differential equation, the focus was on the homogeneous part, given as \[y'' - 5y' + 6y = 0\]To find the characteristic equation, we substitute possible solutions of the form \(y = e^{rt}\) into the equation. This results in a characteristic equation of the form:\[r^2 - 5r + 6 = 0.\]Using the quadratic formula or factoring, we find the roots: \(r = 2\) and \(r = 3\). These roots indicate the exponents in the expression for the complementary solution. They also provide the fundamental solutions that represent the homogeneous system. The complementary solution, \(y_h\), derived from these roots is:\[y_h = C_1 e^{2t} + C_2 e^{3t}.\]

Undetermined Coefficients

The method of undetermined coefficients is a powerful tool used to find particular solutions for non-homogeneous linear differential equations. This method is especially effective for differential equations where the non-homogeneous part (the forcing function) is a simple polynomial, sinusoidal, or exponential function.
For instance, in our example, the non-homogeneous term is \(2e^t\). By the method of undetermined coefficients, we assume a particular solution of the form \(y_p = Ae^t\), where \(A\) is an unknown coefficient to be determined.The key steps include:

• Choosing a trial solution that matches the form of the non-homogeneous term.
• Deriving the derivatives of the trial function.
• Substituting these derivatives into the original differential equation.
• Solving for the unknown coefficients by matching terms.

In our case, this results in \(A = \frac{5}{3}\), giving a particular solution:\[y_p = \frac{5}{3}e^t.\]

Wronskian

The Wronskian is a determinant used in the theory of differential equations to establish whether a set of solutions is linearly independent. In the variation of parameters method, it plays a pivotal role in constructing the particular solution for a linear differential equation.To calculate the Wronskian for the given fundamental solutions \(e^{2t}\) and \(e^{3t}\), we set up the Wronskian determinant as follows:\[W = \begin{vmatrix} e^{2t} & e^{3t} \ 2e^{2t} & 3e^{3t} \end{vmatrix}.\]Upon calculating, we find\[W = 3e^{5t} - 2e^{5t} = e^{5t}.\]This determination is crucial because it divides elements within the integration process when you calculate functions \(u_1\) and \(u_2\) necessary for the particular solution in variation of parameters. It ensures the independence of the solutions involved and assures that our particular solution is valid.

Particular Solution

Finding a particular solution for a non-homogeneous differential equation is an essential step that pinpoints the specific function which satisfies the entire equation. The method of variation of parameters is used here precisely due to its general applicability.For the differential equation\[y'' - 5y' + 6y = 2e^t,\]we seek a function \(y_p\) whose derivatives satisfy the whole equation. We adopt a formula based on the Wronskian and integrals, which involve the functions \(u_1(t)\) and \(u_2(t)\). These functions are solved from:

• \(u_1(t) = \int \frac{-f(t) y_2(t)}{W(t)} \, dt\)
• \(u_2(t) = \int \frac{f(t) y_1(t)}{W(t)} \, dt\)

In this context, \(f(t)\) is the non-homogeneous part \(2e^t\), \(y_1(t) = e^{2t}\), and \(y_2(t) = e^{3t}\).After integration:

• \(u_1(t) = e^{-2t}\)
• \(u_2(t) = \frac{2}{3}e^{-3t}\)

Hence, the particular solution becomes:\[y_p = e^{-2t}e^{2t} + \frac{2}{3}e^{-3t}e^{3t} = 1 + \frac{2}{3} = \frac{5}{3}.\]This particular solution fits when combined with the complementary solution, providing a comprehensive solution to the differential equation.