Question

Match List-I with List-II 

List-I	\(F(x)\)	List-II	Particular Integral Rule
A	\(F(x)=e^{ax}\)	III	\(\dfrac{e^{ax}}{f(a)}, \quad f(a)\neq 0\)
B	\(F(x)=\sin ax\)	II	\(\dfrac{\sin ax}{f(-a^2)}\)
C	\(F(x)=xe^{ax}\)	I	\(\left(\dfrac{x}{f(a)}-\dfrac{f'(a)}{[f(a)]^2}\right)e^{ax}, \quad f(a)\neq 0\)
D	\(F(x)=\cos ax\)	IV	\(D^2\) is replaced by \(-a^2\) in \(f(D)\)

• A. A-I, B-II, C-III, D-IV
• B. A-III, B-IV, C-I, D-II
• C. A-III, B-II, C-I, D-IV
• D. A-I, B-IV, C-III, D-II

Hint:

For \(e^{ax}\), replace \(D\) by \(a\); for \(\sin ax\) and \(\cos ax\), replace \(D^2\) by \(-a^2\).

Answer 1

The Correct Option is C

Concept:

For linear differential equations with constant coefficients: \[ f(D)y = F(x) \] the Particular Integral (P.I.) is obtained using standard operator rules. 

Step 1: For \(F(x)=e^{ax}\)
\[ P.I.=\frac{e^{ax}}{f(a)}, \quad f(a)\neq0 \] Therefore: \[ A \rightarrow III \] 
Step 2: For \(F(x)=\sin ax\)
\[ P.I.=\frac{\sin ax}{f(-a^2)} \] Therefore: \[ B \rightarrow II \] 
Step 3: For \(F(x)=xe^{ax}\)
\[ P.I.=\left(\frac{x}{f(a)}-\frac{f'(a)}{[f(a)]^2}\right)e^{ax} \] Therefore: \[ C \rightarrow I \] 
Step 4: For \(F(x)=\cos ax\)
For sine or cosine functions, replace \(D^2\) by: \[ -a^2 \] in \(f(D)\). Therefore: \[ D \rightarrow IV \] 

Hence, \[ A-III,\quad B-II,\quad C-I,\quad D-IV \] 
\[ \therefore \text{Correct Answer is (C)} \]