Question

Particular integral of $f(D)y = \cos ax$ is

• A. $\frac{1}{f(-a^2)} \cos ax$ iff $f(-a^2) \neq 0$
• B. $\frac{1}{f(a^2)} \cos ax$ iff $f(-a^2) \neq 0$
• C. $\frac{1}{f(a)} \cos ax$ iff $f(-a^2) \neq 0$
• D. $\frac{1}{2} \cos ax$ iff $f(-a^2) \neq 0$

Hint:

Be extremely careful: replace $D^2$ with $-(a^2)$, not $(-a)^2$. For example, if $a=3$, then $D^2$ should be replaced by $-9$, not $+9$. This is the most common error students make in this section of calculus.

Answer 1

The Correct Option is A

1. General Rule for Trigonometric P.I.: For an equation of the form $f(D)y = \cos(ax)$ or $f(D)y = \sin(ax)$, where $f(D)$ is a polynomial in the differential operator $D$, the standard rule is to replace $D^2$ with $-a^2$.

2. Mathematical Expression: The particular integral is given by: $$P.I. = \frac{1}{f(D^2)} \cos ax$$ Applying the substitution $D^2 \to -a^2$: $$P.I. = \frac{1}{f(-a^2)} \cos ax$$

3. The Condition of Validity: This substitution is only valid if the denominator does not become zero. If $f(-a^2) = 0$, the formula results in division by zero, which is undefined. In such a "case of failure," a different approach involving multiplication by $x$ and differentiation of the operator is required. Thus, the formula $\frac{1}{f(-a^2)} \cos ax$ is valid if and only if (iff) $f(-a^2) \neq 0$.