Math Problem Statement

Solve the differential equation y'' + y = sec^2(t) using the method of variation of parameters. Given that the homogeneous solution is y_h = C1 * cos(t) + C2 * sin(t).

Solution

The problem in the image deals with solving a second-order differential equation using the method of variation of parameters. The equation given is:

y+y=sec2(t)y'' + y = \sec^2(t)

The homogeneous solution yhy_h is provided as:

yh=C1cos(t)+C2sin(t)y_h = C_1 \cos(t) + C_2 \sin(t)

Step-by-Step Solution

  1. Solve the homogeneous equation: The homogeneous differential equation corresponding to the problem is: y+y=0y'' + y = 0 The characteristic equation for this is: r2+1=0r^2 + 1 = 0 Solving for rr: r=±ir = \pm i This gives the general solution of the homogeneous equation: yh=C1cos(t)+C2sin(t)y_h = C_1 \cos(t) + C_2 \sin(t) which is already provided.

  2. Variation of Parameters: To solve the non-homogeneous equation using variation of parameters, we assume a particular solution of the form: yp=u1(t)cos(t)+u2(t)sin(t)y_p = u_1(t) \cos(t) + u_2(t) \sin(t) where u1(t)u_1(t) and u2(t)u_2(t) are functions to be determined.

  3. Determine u1(t)u_1(t) and u2(t)u_2(t): We need to solve for u1(t)u_1(t) and u2(t)u_2(t) using the system of equations derived from: u1(t)cos(t)+u2(t)sin(t)=0u_1'(t) \cos(t) + u_2'(t) \sin(t) = 0 and u1(t)sin(t)+u2(t)cos(t)=sec2(t)-u_1'(t) \sin(t) + u_2'(t) \cos(t) = \sec^2(t)

    Solving these equations, we obtain: u1(t)=sec2(t)sin(t)u_1'(t) = -\sec^2(t) \sin(t) u2(t)=sec2(t)cos(t)u_2'(t) = \sec^2(t) \cos(t)

  4. Integrate to find u1(t)u_1(t) and u2(t)u_2(t):

    • Integrating u1(t)u_1'(t): u1(t)=sec2(t)sin(t)dtu_1(t) = -\int \sec^2(t) \sin(t) \, dt This integral can be solved using substitution.

    • Integrating u2(t)u_2'(t): u2(t)=sec2(t)cos(t)dtu_2(t) = \int \sec^2(t) \cos(t) \, dt Similarly, solve this integral.

  5. Construct the particular solution: After finding u1(t)u_1(t) and u2(t)u_2(t), substitute them back into the particular solution yp=u1(t)cos(t)+u2(t)sin(t)y_p = u_1(t) \cos(t) + u_2(t) \sin(t).

  6. General solution: The general solution to the differential equation is: y(t)=yh+yp=C1cos(t)+C2sin(t)+ypy(t) = y_h + y_p = C_1 \cos(t) + C_2 \sin(t) + y_p

Would you like more detailed steps on the integration process or any specific part of the solution?

Follow-up Questions:

  1. How do we solve the integrals in the variation of parameters method?
  2. What is the significance of the homogeneous solution in solving differential equations?
  3. Can variation of parameters be used for any non-homogeneous differential equation?
  4. How do we verify that the solution satisfies the original differential equation?
  5. What are the alternative methods for solving second-order differential equations?

Tip:

When solving differential equations, always check if the particular solution you found is valid by substituting it back into the original equation.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Variation of Parameters
Second-Order Differential Equations

Formulas

Homogeneous solution: y_h = C1 * cos(t) + C2 * sin(t)
Particular solution: y_p = u1(t) * cos(t) + u2(t) * sin(t)
System of equations for u1'(t) and u2'(t): u1'(t) * cos(t) + u2'(t) * sin(t) = 0 and -u1'(t) * sin(t) + u2'(t) * cos(t) = sec^2(t)

Theorems

Variation of Parameters

Suitable Grade Level

Undergraduate Mathematics