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In  question 24 why we got f(a) =0 and can you explain it to me how we calculate f(a) in solution

In Question 24, the differential equation given is \((D^2 - 1)y = \sinh x\).

### Step-by-Step Explanation:

1. **Given Differential Equation:**
   \[
   (D^2 - 1)y = \sinh x \tag{1}
   \]
   Here, \(D\) represents the differential operator, \(\frac{d}{dx}\).

2. **Non-Homogeneous Linear Differential Equation Form:**
   A non-homogeneous linear differential equation can be written as:
   \[
   f(D)y = X \tag{2}
   \]
   Comparing (1) and (2), we get:
   \[
   f(D) = D^2 - 1 \quad \text{and} \quad X = \sinh x
   \]

3. **General Solution:**
   The general solution \(y\) of a non-homogeneous linear differential equation is given by:
   \[
   y = C.F. + P.I. \tag{3}
   \]
   where \(C.F.\) is the complementary function and \(P.I.\) is the particular integral.

4. **Auxiliary Equation:**
   To find the \(C.F.\), solve the auxiliary equation:
   \[
   f(m) = 0
   \]
   For \(f(D) = D^2 - 1\):
   \[
   m^2 - 1 = 0
   \]
   Solving this, we get:
   \[
   (m - 1)(m + 1) = 0 \implies m_1 = 1, \, m_2 = -1
   \]
   Hence, the roots are real and distinct. Therefore, the complementary function is:
   \[
   C.F. = c_1 e^x + c_2 e^{-x}
   \]

5. **Particular Integral:**
   The particular integral (P.I.) is found using:
   \[
   P.I. = \frac{X}{f(D)}
   \]
   For \(X = \sinh x\):
   \[
   P.I. = \frac{\sinh x}{D^2 - 1}
   \]
   Now, \(\sinh x\) can be written as a combination of exponential functions:
   \[
   \sinh x = \frac{e^x - e^{-x}}{2}
   \]
   Thus:
   \[
   P.I. = \frac{\frac{e^x - e^{-x}}{2}}{D^2 - 1} = \frac{e^x}{2(D^2 - 1)} - \frac{e^{-x}}{2(D^2 - 1)}
   \]

6. **Calculation of \(f(a)\):**
   For \(e^{ax}\) as a particular solution, we need to evaluate \(f(a)\) where \(D = a\):
   \[
   f(a) = a^2 - 1
   \]
   - When \(a = 1\):
     \[
     f(1) = 1^2 - 1 = 0
     \]
   - When \(a = -1\):
     \[
     f(-1) = (-1)^2 - 1 = 0
     \]

7. **Formulating the Particular Integral:**
   When \(f(a) = 0\), this indicates resonance, and we need to use a modified approach to find the particular integral. For \(\sinh x = \frac{e^x - e^{-x}}{2}\), the resonant terms are \(e^x\) and \(e^{-x}\). Therefore, we adjust the P.I. to:
   \[
   P.I. = \frac{x e^x}{2(D-1)} - \frac{x e^{-x}}{2(D+1)}
   \]
   Substituting:
   \[
   P.I. = \frac{x e^x}{2(1-1)} - \frac{x e^{-x}}{2(-1+1)} \rightarrow \frac{x e^x}{2 \cdot 0} - \frac{x e^{-x}}{2 \cdot 0}
   \]
   This approach reveals the need to handle terms correctly, often requiring further steps or techniques like undetermined coefficients.

8. **Final Solution:**
   After finding \(P.I.\), the general solution combines \(C.F.\) and \(P.I.\):
   \[
   y = c_1 e^x + c_2 e^{-x} + P.I.
   \]

Would you like any specific step further detailed or have any more questions on this topic?

Here are some questions you might ask next:
1. How do you find the complementary function (C.F.) in a differential equation?
2. What is the significance of resonance in differential equations?
3. How do you handle repeated roots in the auxiliary equation?
4. What is the method of undetermined coefficients?
5. How do you solve non-homogeneous differential equations using the variation of parameters?

**Tip:** Always check the roots of the auxiliary equation carefully as they dictate the form of the complementary function and can affect the approach to finding the particular integral.

Learn how to solve differential equations using the method of finding f(a) = 0, illustrated with Question 24's example. This problem covers non-homogeneous linear differential equations, particular integrals, and formulas like f(D) = D^2 - 1. Suitable for undergraduate students studying differential equations.

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