To solve the differential equation 1 4y” + 3y’ = x^2 by the method of undetermined coefficients, we first rewrite it in standard form:
4y” + 3y’ = x^2
Next, we need to find the complementary solution (the solution to the homogeneous equation) and the particular solution (the solution to the non-homogeneous equation).
Step 1: Find the Complementary Solution
The homogeneous part of the equation is:
4y” + 3y’ = 0
To solve this, we assume a solution of the form y = e^{rt}, substituting it into the homogeneous equation leads us to the characteristic equation:
4r^2 + 3r = 0
Factoring out r gives:
r(4r + 3) = 0
This gives us r = 0 or 4r + 3 = 0 → r = -3/4. So, the complementary solution is:
y_c = C_1 + C_2 e^{-3x/4}
Step 2: Find the Particular Solution
For the particular solution y_p, since the right side of the equation is x^2, we can assume a polynomial form:
y_p = Ax^2 + Bx + C
We need to find the first and second derivatives:
y_p’ = 2Ax + B
y_p” = 2A
Substituting y_p, y_p’, and y_p” back into the original equation:
4(2A) + 3(2Ax + B) = x^2
This simplifies to:
8A + (6A)x + 3B = x^2
Comparing coefficients, we need:
- Coefficient of x²: 6A = 1 → A = 1/6
- Coefficient of x: 3B = 0 → B = 0
- No constant term: 8A = 0 + C → C = 0
Thus, we have:
y_p = (1/6)x^2
Step 3: Combine Solutions
The general solution of the differential equation is the sum of the complementary and particular solutions:
y = y_c + y_p = C_1 + C_2 e^{-3x/4} + (1/6)x^2
This provides us with the complete solution to the differential equation.