To solve the differential equation y” + 14y’ + 4y = e^{2x} using the method of undetermined coefficients, we first need to find the complementary solution (homogeneous solution) and then find the particular solution.
Step 1: Find the complementary solution
The corresponding homogeneous equation is:
y” + 14y’ + 4y = 0
To solve this, we assume a solution of the form y = e^{rx}. Substituting this into the homogeneous equation gives us the characteristic equation:
r² + 14r + 4 = 0
Using the quadratic formula, r = rac{-b \pm \sqrt{b^2 – 4ac}}{2a}, where a=1, b=14, and c=4, we find:
r = \frac{-14 \pm \sqrt{14^2 – 4\cdot1\cdot4}}{2\cdot1} = \frac{-14 \pm \sqrt{196 – 16}}{2} = \frac{-14 \pm \sqrt{180}}{2} = \frac{-14 \pm 6\sqrt{5}}{2}
This gives us two distinct real roots:
r_1 = -7 + 3\sqrt{5}, \, r_2 = -7 – 3\sqrt{5}
Thus, the complementary solution is:
y_c = C_1 e^{(-7 + 3\sqrt{5})x} + C_2 e^{(-7 – 3\sqrt{5})x}
Step 2: Find the particular solution
Now, we look for a particular solution using the method of undetermined coefficients. Since the right-hand side is e^{2x}, we assume a particular solution of the form:
y_p = Ae^{2x}
Next, we find the first and second derivatives:
y_p’ = 2Ae^{2x}, \, y_p” = 4Ae^{2x}
We substitute y_p, y_p’, and y_p” into the original differential equation:
4Ae^{2x} + 14(2Ae^{2x}) + 4(Ae^{2x}) = e^{2x}
This simplifies to:
(4A + 28A + 4A)e^{2x} = e^{2x}
36Ae^{2x} = e^{2x}
Setting the coefficients equal gives:
36A = 1 \Rightarrow A = \frac{1}{36}
Thus, the particular solution is:
y_p = \frac{1}{36}e^{2x}
Step 3: General solution
The general solution to the differential equation is the sum of the complementary and particular solutions:
y = y_c + y_p = C_1 e^{(-7 + 3\sqrt{5})x} + C_2 e^{(-7 – 3\sqrt{5})x} + \frac{1}{36}e^{2x}
This is the solution to the given differential equation.