For a linear differential equation with operator f(D), the particular integral for Q(x) = e^(ax) is given by:
A
1/f(D) · e^(ax) = e^(ax)/f(a)
B
1/f(D) · e^(ax) = e^(ax)/f(D)
C
1/f(D) · e^(ax) = e^(ax)/a
D
1/f(D) · e^(ax) = f(a)
Analysis & Theory
If Q(x) = e^(ax), then PI = e^(ax)/f(a), provided f(a) ≠ 0.
If f(a) = 0 for Q(x) = e^(ax), then the PI is obtained by:
A
Multiplying by x and dividing by f’(a)
B
Dividing by a²
C
Using separation of variables
D
Integrating factor method
Analysis & Theory
When f(a) = 0, PI = x·e^(ax)/f’(a).
For Q(x) = sin(ax) or cos(ax), the particular integral is:
A
1/f(D² + a²) · [sin(ax) or cos(ax)]
B
1/f(D + a²) · [sin(ax) or cos(ax)]
C
1/f(D) · [sin(ax) or cos(ax)]
D
1/a · [sin(ax) or cos(ax)]
Analysis & Theory
We use D² → -a². Hence PI = 1/f(-a²) [sin(ax) or cos(ax)], if f(-a²) ≠ 0.
If f(-a²) = 0 for Q(x) = sin(ax) or cos(ax), then:
A
Multiply by x and divide by derivative of f(D²)
B
No solution exists
C
It becomes homogeneous
D
Change substitution y = vx
Analysis & Theory
When f(-a²) = 0, we multiply by x and use derivative f’(-a²) in denominator.
For equation (D² + 4)y = sin(2x), the particular integral is:
A
PI = sin(2x)/f(-4)
B
PI = cos(2x)/f(-2)
C
PI = sin(2x)/0
D
PI = (1/4)sin(2x)
Analysis & Theory
Here f(D²) = D² + 4 ⇒ f(-a²) = (-4)+4=0. Since denominator vanishes, we use PI = (x/2)cos(2x).
For equation (D² + 1)y = cos(x), the PI is:
A
PI = cos(x)/f(-1)
B
PI = cos(x)/2
C
PI = cos(x)/0
D
PI = (1/2)x sin(x)
Analysis & Theory
Here f(D²) = D² + 1 ⇒ f(-1) = 0 ⇒ multiply by x. PI = (x/2)sin(x).
For (D - 2)y = e^(2x), the PI is:
A
PI = e^(2x)/(2-2)
B
PI = x e^(2x)
C
PI = e^(2x)/(D-2)
D
PI = Ce^(2x)
Analysis & Theory
Since f(a) = (2-2)=0, PI = x e^(2x).
In operator method, the rule for exponential function Q(x) = e^(ax) is:
A
Replace D by a
B
Replace D by -a²
C
Replace D² by a²
D
No substitution needed
Analysis & Theory
We replace D by a in f(D) when Q(x) = e^(ax).
In operator method, the rule for trigonometric functions Q(x) = sin(ax), cos(ax) is:
A
Replace D by a
B
Replace D² by -a²
C
Replace D by -a
D
Replace D² by a²
Analysis & Theory
We replace D² by -a² for sin(ax) and cos(ax).
The complete solution of a non-homogeneous equation is:
A
Particular integral only
B
Complementary function only
C
CF + PI
D
Depends on Q(x)
Analysis & Theory
General solution = Complementary Function (CF) + Particular Integral (PI).