Find a particular solution of the following equation using the Method of Undetermined Coefficients. y''-y'-2y=6x+7

The given second-order linear differential equation is:

\[
y'' - y' - 2y = 6x + 7
\]

To find a particular solution using the **Method of Undetermined Coefficients**, follow these steps:

### Step 1: Solve the homogeneous equation.
The corresponding homogeneous equation is:

\[
y'' - y' - 2y = 0
\]

The characteristic equation for this is:

\[
r^2 - r - 2 = 0
\]

Solve for \( r \) using the quadratic formula:

\[
r = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)} = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2}
\]

Thus, \( r = 2 \) and \( r = -1 \). Therefore, the general solution to the homogeneous equation is:

\[
y_h(x) = C_1 e^{2x} + C_2 e^{-x}
\]

### Step 2: Guess the form of the particular solution.
The right-hand side of the original equation is \( 6x + 7 \), which is a first-degree polynomial. When the non-homogeneous term is a polynomial, the guess for the particular solution is a polynomial of the same degree. Hence, assume:

\[
y_p(x) = Ax + B
\]

### Step 3: Find the derivatives of \( y_p(x) \).
Compute the first and second derivatives of the assumed particular solution:

\[
y_p'(x) = A
\]
\[
y_p''(x) = 0
\]

### Step 4: Substitute \( y_p(x) \), \( y_p'(x) \), and \( y_p''(x) \) into the original equation.
Substitute these into the differential equation \( y'' - y' - 2y = 6x + 7 \):

\[
0 - A - 2(Ax + B) = 6x + 7
\]

Simplify the left-hand side:

\[
-A - 2Ax - 2B = 6x + 7
\]

Now, match the coefficients of like terms on both sides:

1. For \( x \): \(-2A = 6 \), so \( A = -3 \).
2. For the constant term: \(-A - 2B = 7\), so \(-(-3) - 2B = 7\), which simplifies to \( 3 - 2B = 7 \), or \( -2B = 4 \), so \( B = -2 \).

### Step 5: Write the particular solution.
Thus, the particular solution is:

\[
y_p(x) = -3x - 2
\]

### Step 6: Write the general solution.
The general solution to the differential equation is the sum of the homogeneous and particular solutions:

\[
y(x) = y_h(x) + y_p(x) = C_1 e^{2x} + C_2 e^{-x} - 3x - 2
\]

This is the general solution to the given differential equation.

Would you like further details or explanations? Feel free to ask!

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Here are five related questions to expand your understanding:

1. How does the Method of Undetermined Coefficients work for trigonometric functions on the right-hand side?
2. Can the characteristic equation have complex roots, and how does that affect the solution?
3. How do you solve a differential equation with a non-polynomial forcing function, such as \( e^x \)?
4. What happens if the particular solution overlaps with the homogeneous solution?
5. How can you verify the solution of a differential equation?

**Tip**: Always substitute your particular solution back into the original equation to verify correctness.

Find a particular solution of the following equation using the Method of Undetermined Coefficients. y'' - y' - 2y = 6x + 7

Learn how to solve the second-order linear differential equation y'' - y' - 2y = 6x + 7 using the Method of Undetermined Coefficients. This step-by-step guide includes solving the homogeneous equation, guessing the form of the particular solution, and calculating constants. Suitable for students studying undergraduate differential equations.

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