Question

Undetermined coefficients Use the Method of Undetermined Coefficients to solve the following differential equations.

1) y′′ + y′ + 5y = 3x − 2

2) y′′ + 6y′ + 9y = 4e−³ˣ, y(0) = 2, y′ (0) = 5

3) 4y′′ − 4y′ − 3y = cos(2x)

4) 1/4 y′′ + y′ + y = x + 2 + eˣ

Answer 1

To solve differential equations using the Method of Undetermined Coefficients, solve the characteristic equation for the homogeneous solution and propose a particular solution based on the non-homogeneous term (polynomial, exponential, or trigonometric).

Step-by-step explanation:

To solve the given differential equations using the Method of Undetermined Coefficients, one must first determine the homogeneous solution of the differential equation and then propose a suitable form for the particular solution based on the type of non-homogeneous term.

1. For the equation y′′ + y′ + 5y = 3x − 2, the characteristic equation is r² + r + 5 = 0. Solving the characteristic equation will give the homogeneous solution. The particular solution can be proposed based on the polynomial on the right side.

2. For y′′ + 6y′ + 9y = 4e⁻³ˣ, with initial conditions y(0) = 2 and y′ (0) = 5, a characteristic equation is formed and the initial conditions are used to find the constants of the homogeneous solution. The particular solution will involve a term with an exponential function due to e⁻³ˣ on the right side.

3. The differential equation 4y′′ − 4y′ − 3y = cos(2x) also requires finding the characteristic equation's solution for the homogeneous part. The particular solution is proposed based on the trigonometric function cosine.

4. Finally, for 1/4 y′′ + y′ + y = x + 2 + eˣ, the characteristic equation provides the homogeneous solution, while the particular solution is likely a combination of a polynomial and an exponential term corresponding to x + 2 + eˣ.

Answer 2

The Method of Undetermined Coefficients is used to solve differential equations by finding a particular solution. Various forms of the particular solution are used depending on the type of differential equation. By substituting the particular solution into the differential equation, we can solve for the unknown coefficients.

Step-by-step explanation:

1) y′′ + y′ + 5y = 3x − 2:

The particular solution can be of the form yp = Ax + B. By substituting yp into the differential equation, we can solve for the constants A and B.

2) y′′ + 6y′ + 9y = 4e−³ˣ, y(0) = 2, y′ (0) = 5:

The particular solution can be of the form yp = Ae^(-3x). By substituting yp into the differential equation and applying the initial conditions, we can solve for the constant A.

3) 4y′′ − 4y′ − 3y = cos(2x):

The particular solution can be of the form yp = A*cos(2x) + B*sin(2x). By substituting yp into the differential equation, we can solve for the constants A and B.

4) 1/4 y′′ + y′ + y = x + 2 + eˣ:

The particular solution can be of the form yp = Ax + B + Ce^x. By substituting yp into the differential equation, we can solve for the constants A, B, and C.