y”-3y’-4y=x+2
To solve the differential equation \( y'' - 3y' - 4y = x + 2 \), we can use the method of undetermined coefficients, as it's a linear non-homogeneous ordinary differential equation. This involves two main steps: finding the complementary solution to the associated homogeneous equation and a particular solution to the non-homogeneous equation.
Step 1: Solve the homogeneous equation
The associated homogeneous equation is:
\[ y'' - 3y' - 4y = 0 \]
To solve this, we first find the characteristic equation by assuming a solution of the form \( y = e^{rt} \):
\[ r^2 - 3r - 4 = 0 \]
Now, we factor or use the quadratic formula to solve for \( r \):
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \] \[ = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2} \]
This gives us the roots:
\[ r_1 = 4 \quad \text{and} \quad r_2 = -1 \]
Thus, the complementary solution \( y_c \) is:
\[ y_c = C_1 e^{4x} + C_2 e^{-x} \]
where \( C_1 \) and \( C_2 \) are constants.
Step 2: Find a particular solution
Next, we need to find a particular solution \( y_p \) for the non-homogeneous part \( x + 2 \). We can guess a polynomial form for the particular solution since the right-hand side is a polynomial. Therefore, we can try:
\[ y_p = Ax + B \]
where \( A \) and \( B \) are constants to be determined. We need to compute \( y' \) and \( y'' \):
\[ y' = A \] \[ y'' = 0 \]
Substituting \( y_p \) into the left-hand side of the original equation gives:
\[ 0 - 3A - 4(Ax + B) = x + 2 \]
This simplifies to:
\[ -4Ax - (3A + 4B) = x + 2 \]
Now, we equate the coefficients from both sides:
For \( x \): \[ -4A = 1 \implies A = -\frac{1}{4} \]
For the constant term: \[ -3A - 4B = 2 \] Substituting \( A = -\frac{1}{4} \): \[ -3\left(-\frac{1}{4}\right) - 4B = 2 \] \[ \frac{3}{4} - 4B = 2 \] Rearranging gives: \[ -4B = 2 - \frac{3}{4} = \frac{8}{4} - \frac{3}{4} = \frac{5}{4} \] \[ B = -\frac{5}{16} \]
Thus, the particular solution is:
\[ y_p = -\frac{1}{4}x - \frac{5}{16} \]
Step 3: Write the general solution
The general solution \( y \) is the sum of the complementary and particular solutions:
\[ y = y_c + y_p = C_1 e^{4x} + C_2 e^{-x} - \frac{1}{4}x - \frac{5}{16} \]
Final Result
The general solution to the differential equation is:
\[ y = C_1 e^{4x} + C_2 e^{-x} - \frac{1}{4}x - \frac{5}{16} \]
where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial or boundary conditions.
