How to Solve the Differential Equation 14 y” + y’ + 4y = x^2?

To solve the differential equation 14 y” + y’ + 4y = x^2 using the method of undetermined coefficients, we first need to determine the complementary solution and a particular solution.

Step 1: Find the Complementary Solution (y_c)

The characteristic equation is derived from the homogeneous part of the differential equation:

14r² + r + 4 = 0

Using the quadratic formula, r = (-b ± √(b² - 4ac)) / 2a, where a = 14, b = 1, and c = 4, we compute:

r = ( -1 ± √(1² – 4 * 14 * 4) ) / (2 * 14)

This simplifies to:

r = ( -1 ± √(-223) ) / 28

Since the discriminant is negative, we have complex roots:

r = -1/28 ± (√223 / 28)i

The complementary solution is then given by:

y_c = e^(-1/28)x (C₁ cos(√223/28 * x) + C₂ sin(√223/28 * x))

Step 2: Find the Particular Solution (y_p)

Next, we find a particular solution by assuming a form that fits the non-homogeneous part x². Since the degree is 2, we take:

y_p = Ax² + Bx + C

Now we need the first and second derivatives:

y_p‘ = 2Ax + B

y_p” = 2A

Substituting y_p, y_p‘, and y_p” into the differential equation:

14(2A) + (2Ax + B) + 4(Ax² + Bx + C) = x²

Grouping like terms:

(14*2A + 4A)x² + (2A + 4B)x + (14*2A + B + 4C) = x²

This yields the system of equations:

• 4A + 28A = 1 (Coefficient of x²)
• 2A + 4B = 0 (Coefficient of x)
• 14*2A + B + 4C = 0 (Constant term)

Simplifying these equations gives:

• 32A = 1 ⇒ A = 1/32
• 2(1/32) + 4B = 0 ⇒ B = -1/64
• 28(1/32) – 1/64 + 4C = 0 ⇒ C = -7/64

Thus, the particular solution is:

y_p = (1/32)x² – (1/64)x – (7/64)

Step 3: General Solution

The general solution is the sum of the complementary and particular solutions:

y = y_c + y_p

y = e^(-1/28)x (C₁ cos(√223/28 * x) + C₂ sin(√223/28 * x)) + (1/32)x² – (1/64)x – (7/64)

This completes the solution of the differential equation using the method of undetermined coefficients.