Math Problem Statement

Find particular integral

Find the particular integral for the differential equation \( \frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} + \frac{dy}{dx} = e^{2x} + \sin(3x) \).

Solution

The given differential equation is:

d3ydx3+2d2ydx2+dydx=e2x+sin(3x)\frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} + \frac{dy}{dx} = e^{2x} + \sin(3x)

To find the particular integral, we can use the method of undetermined coefficients or the annihilator method. The right-hand side consists of two terms: e2xe^{2x} and sin(3x)\sin(3x), so the particular integral will be found for each of these separately.

Step-by-step solution:

  1. Solve for the particular integral of e2xe^{2x}:

    • The corresponding homogeneous equation is: D3y+2D2y+Dy=0D^3 y + 2D^2 y + D y = 0 where DD represents differentiation with respect to xx.
    • The right-hand side e2xe^{2x} suggests that we try a solution of the form yp=Ae2xy_p = A e^{2x}.
    • Substituting this into the left-hand side of the equation: (D3+2D2+D)(Ae2x)=A(8+8+2)e2x=18Ae2x(D^3 + 2D^2 + D)(A e^{2x}) = A(8 + 8 + 2)e^{2x} = 18A e^{2x}
      • Matching coefficients with the right-hand side e2xe^{2x}, we get: 18A=1    A=11818A = 1 \implies A = \frac{1}{18}
      • Therefore, the particular integral for the e2xe^{2x} term is: yp1=118e2xy_{p1} = \frac{1}{18}e^{2x}
  2. Solve for the particular integral of sin(3x)\sin(3x):

    • The term sin(3x)\sin(3x) suggests we try a solution of the form: yp2=Bcos(3x)+Csin(3x)y_{p2} = B \cos(3x) + C \sin(3x)

    • Substituting into the left-hand side: (D3+2D2+D)(Bcos(3x)+Csin(3x))=(27Bcos(3x)27Csin(3x))+(18Bcos(3x)18Csin(3x))+(3Bsin(3x)+3Ccos(3x))(D^3 + 2D^2 + D)(B \cos(3x) + C \sin(3x)) = (-27B \cos(3x) - 27C \sin(3x)) + (-18B \cos(3x) - 18C \sin(3x)) + (-3B \sin(3x) + 3C \cos(3x)) Grouping the terms for cos(3x)\cos(3x) and sin(3x)\sin(3x), we get: cos(3x):(27B18B+3C)=0andsin(3x):(27C18C3B)=1\cos(3x) : (-27B - 18B + 3C) = 0 \quad \text{and} \quad \sin(3x) : (-27C - 18C - 3B) = 1 Solving these two equations:

      • 45B+3C=0    C=15B-45B + 3C = 0 \implies C = 15B
      • 45C3B=1    45(15B)3B=1    675B3B=1    678B=1    B=1678-45C - 3B = 1 \implies -45(15B) - 3B = 1 \implies -675B - 3B = 1 \implies -678B = 1 \implies B = -\frac{1}{678}
      • C=15B=15678=5226C = 15B = -\frac{15}{678} = -\frac{5}{226}

      Therefore, the particular integral for the sin(3x)\sin(3x) term is: yp2=1678cos(3x)5226sin(3x)y_{p2} = -\frac{1}{678}\cos(3x) - \frac{5}{226}\sin(3x)

Final Particular Integral:

yp=yp1+yp2=118e2x1678cos(3x)5226sin(3x)y_p = y_{p1} + y_{p2} = \frac{1}{18} e^{2x} - \frac{1}{678}\cos(3x) - \frac{5}{226}\sin(3x)

This is the particular integral for the given differential equation.

Let me know if you'd like any further details or have any questions!

5 Related Questions:

  1. How do we find the complementary function for the given differential equation?
  2. Can you explain the method of undetermined coefficients in more detail?
  3. How do we apply the annihilator method to this type of problem?
  4. What is the difference between homogeneous and non-homogeneous differential equations?
  5. How would the solution change if the right-hand side were e2x+cos(3x)e^{2x} + \cos(3x) instead of sin(3x)\sin(3x)?

Tip:

For non-homogeneous linear differential equations, always check the form of the non-homogeneous term to choose an appropriate trial solution for the particular integral.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Particular Integral
Method of Undetermined Coefficients

Formulas

\(D^3 y + 2D^2 y + D y = 0\) (Homogeneous equation)
Particular Integral for \(e^{2x}\) term: \(y_{p1} = \frac{1}{18}e^{2x}\)
Particular Integral for \(\sin(3x)\) term: \(y_{p2} = -\frac{1}{678}\cos(3x) - \frac{5}{226}\sin(3x)\)

Theorems

Annihilator Method
Method of Undetermined Coefficients

Suitable Grade Level

Undergraduate - Advanced Mathematics