Math Problem Statement

solve

Use the variation of parameters method to solve the nonhomogeneous equation y'' + 9y = 3 sec(3x).

Solution

To solve the nonhomogeneous equation y+9y=3sec(3x)y'' + 9y = 3 \sec(3x) using the variation of parameters method, we proceed as follows:


Step 1: Solve the corresponding homogeneous equation

The homogeneous equation is: yh+9yh=0.y_h'' + 9y_h = 0.

This is a second-order differential equation with constant coefficients. Assume a solution of the form yh=erxy_h = e^{rx}. Substituting into the homogeneous equation gives: r2+9=0.r^2 + 9 = 0.

Solve for rr: r=±3i.r = \pm 3i.

Thus, the solution to the homogeneous equation is: yh=c1cos(3x)+c2sin(3x),y_h = c_1 \cos(3x) + c_2 \sin(3x), where c1c_1 and c2c_2 are arbitrary constants.


Step 2: Formulate the variation of parameters solution

The particular solution ypy_p is given by: yp=u1(x)cos(3x)+u2(x)sin(3x),y_p = u_1(x) \cos(3x) + u_2(x) \sin(3x), where u1(x)u_1(x) and u2(x)u_2(x) are functions to be determined.

Step 3: Derive the equations for u1u_1 and u2u_2

The method of variation of parameters requires: u1(x)cos(3x)+u2(x)sin(3x)=0,u_1'(x) \cos(3x) + u_2'(x) \sin(3x) = 0, 3u1(x)sin(3x)+3u2(x)cos(3x)=3sec(3x).-3u_1'(x) \sin(3x) + 3u_2'(x) \cos(3x) = 3\sec(3x).


Step 4: Solve for u1(x)u_1'(x) and u2(x)u_2'(x)

The system of equations can be expressed as:

\cos(3x) & \sin(3x) \\ -3\sin(3x) & 3\cos(3x) \end{bmatrix} \begin{bmatrix} u_1'(x) \\ u_2'(x) \end{bmatrix} = \begin{bmatrix} 0 \\ 3\sec(3x) \end{bmatrix}.$$ Compute the determinant of the coefficient matrix: $$\Delta = \cos(3x)(3\cos(3x)) - (-3\sin(3x))\sin(3x) = 3(\cos^2(3x) + \sin^2(3x)) = 3.$$ Thus, $$\Delta = 3$$. The solutions for $$u_1'(x)$$ and $$u_2'(x)$$ are: $$u_1'(x) = \frac{1}{\Delta} \begin{vmatrix} 0 & \sin(3x) \\ 3\sec(3x) & 3\cos(3x) \end{vmatrix} = \frac{1}{3} \left( 0 - 3\sec(3x)\sin(3x) \right) = -\sec(3x)\sin(3x),$$ $$u_2'(x) = \frac{1}{\Delta} \begin{vmatrix} \cos(3x) & 0 \\ -3\sin(3x) & 3\sec(3x) \end{vmatrix} = \frac{1}{3} \left( 3\sec(3x)\cos(3x) - 0 \right) = \sec(3x)\cos(3x).$$ --- ### Step 5: Integrate to find $$u_1(x)$$ and $$u_2(x)$$ For $$u_1(x)$$: $$u_1(x) = \int -\sec(3x)\sin(3x) \, dx.$$ Let $$u = 3x$$, then $$du = 3dx$$. The integral becomes: $$u_1(x) = -\frac{1}{3} \int \sec(u)\sin(u) \, du = -\frac{1}{3} \ln|\sec(u) + \tan(u)| + C_1.$$ For $$u_2(x)$$: $$u_2(x) = \int \sec(3x)\cos(3x) \, dx.$$ Let $$u = 3x$$, then $$du = 3dx$$. The integral becomes: $$u_2(x) = \frac{1}{3} \int \sec(u)\cos(u) \, du = \frac{1}{3} \ln|\sec(u) + \tan(u)| + C_2.$$ --- ### Step 6: Write the general solution The general solution is: $$y = y_h + y_p = c_1 \cos(3x) + c_2 \sin(3x) + \left( u_1(x) \cos(3x) + u_2(x) \sin(3x) \right).$$ Simplify $$u_1(x)$$ and $$u_2(x)$$ to obtain the full solution. --- Would you like a detailed expansion of the integration steps or the final simplified expression? Let me know! --- ### Related Questions: 1. What is the general method to find the complementary solution for higher-order linear differential equations? 2. How does the variation of parameters method differ from undetermined coefficients? 3. Can variation of parameters be used for non-constant coefficient differential equations? 4. What are some limitations of the variation of parameters method? 5. How do trigonometric identities help simplify solutions involving secant and tangent? --- ### Tip: Always verify your general solution by substituting it back into the original differential equation to check correctness.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Variation of Parameters Method
Trigonometric Functions

Formulas

General solution of nonhomogeneous equations: y = y_h + y_p
Variation of parameters: y_p = u_1(x)y_1(x) + u_2(x)y_2(x)

Theorems

Superposition principle for linear differential equations
Determinant method for solving systems of equations

Suitable Grade Level

Undergraduate Mathematics