Case II: Particular Integral When \( Q(x) = \sin(ax + b) \) or \( \cos(ax + b) \)
Consider the differential equation:
\[
f(D)y = Q(x)
\]
where \( Q(x) = \sin(ax + b) \) or \( \cos(ax + b) \).
The particular integral (P.I.) is given by:
\[
\text{P.I.} = \frac{1}{f(D)} Q(x) = \frac{1}{f(D)} \sin(ax + b) \quad \text{or} \quad \frac{1}{f(D)} \cos(ax + b)
\]
To evaluate this, substitute \( D^2 = -a^2 \):
\[
\text{P.I.} = \frac{1}{f(-a^2)} \sin(ax + b) \quad \text{or} \quad \frac{1}{f(-a^2)} \cos(ax + b)
\]
This is valid provided \( f(-a^2) \neq 0 \).
If \( f(-a^2) = 0 \), then:
\[
\text{P.I.} = x \frac{1}{f'(D)} \sin(ax + b) \quad \text{or} \quad x \frac{1}{f'(D)} \cos(ax + b)
\]
Substitute \( D^2 = -a^2 \):
\[
\text{P.I.} = x \frac{1}{f'(-a^2)} \sin(ax + b) \quad \text{or} \quad x \frac{1}{f'(-a^2)} \cos(ax + b)
\]
This is valid provided \( f'(-a^2) \neq 0 \).
If \( f'(-a^2) = 0 \), then:
\[
\text{P.I.} = x^2 \frac{1}{f''(D)} \sin(ax + b) \quad \text{or} \quad x^2 \frac{1}{f''(D)} \cos(ax + b)
\]
Substitute \( D^2 = -a^2 \):
\[
\text{P.I.} = x^2 \frac{1}{f''(-a^2)} \sin(ax + b) \quad \text{or} \quad x^2 \frac{1}{f''(-a^2)} \cos(ax + b)
\]
This is valid provided \( f''(-a^2) \neq 0 \).
If \( f''(-a^2) = 0 \), continue the same process by increasing the power of \( x \) and differentiating \( f(D) \) further until a non-zero denominator is obtained.
1. Solve the differential equation: \( (D^4 + 10D^2 + 9)y = \cos(2x + 3) \)
Solution:
The given equation is in the form \( f(D)y = Q(x) \), where:
\[
f(D) = D^4 + 10D^2 + 9 \quad \text{and} \quad Q(x) = \cos(2x + 3)
\]
The auxiliary equation is:
\[
f(m) = 0 \Rightarrow m^4 + 10m^2 + 9 = 0 \Rightarrow m = \pm i, \pm 3i
\]
Thus, the complementary function (C.F.) is:
\[
\text{C.F.} = (c_1 \cos x + c_2 \sin x) + (c_3 \cos 3x + c_4 \sin 3x)
\]
The particular integral (P.I.) is:
\[
\text{P.I.} = \frac{1}{D^4 + 10D^2 + 9} \cos(2x + 3)
\]
Substitute \( D^2 = -a^2 = -4 \) (since \( a = 2 \)):
\[
\text{P.I.} = \frac{1}{(-4)^2 + 10(-4) + 9} \cos(2x + 3) = \frac{1}{16 - 40 + 9} \cos(2x + 3) = -\frac{1}{15} \cos(2x + 3)
\]
The complete solution is:
\[
y = \text{C.F.} + \text{P.I.} = (c_1 \cos x + c_2 \sin x) + (c_3 \cos 3x + c_4 \sin 3x) - \frac{1}{15} \cos(2x + 3)
\]
2. Solve the differential equation: \( \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} + y = e^{2x} - \cos^2 x \)
Solution:
The given equation can be written as:
\[
(D^2 + 2D + 1)y = e^{2x} - \cos^2 x
\]
This is in the form \( f(D)y = Q(x) \), where:
\[
f(D) = D^2 + 2D + 1 \quad \text{and} \quad Q(x) = e^{2x} - \cos^2 x
\]
The auxiliary equation is:
\[
f(m) = 0 \Rightarrow m^2 + 2m + 1 = 0 \Rightarrow m = -1, -1
\]
Thus, the complementary function (C.F.) is:
\[
\text{C.F.} = (c_1 + c_2 x) e^{-x}
\]
The particular integral (P.I.) is:
\[
\text{P.I.} = \frac{1}{D^2 + 2D + 1} \left( e^{2x} - \cos^2 x \right)
\]
Simplify \( \cos^2 x \) using the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \):
\[
\text{P.I.} = \frac{1}{D^2 + 2D + 1} e^{2x} - \frac{1}{D^2 + 2D + 1} \left( \frac{1 + \cos 2x}{2} \right)
\]
Evaluate each term separately:
\[
\frac{1}{D^2 + 2D + 1} e^{2x} = \frac{1}{(2)^2 + 2(2) + 1} e^{2x} = \frac{1}{9} e^{2x}
\]
\[
\frac{1}{D^2 + 2D + 1} \left( \frac{1 + \cos 2x}{2} \right) = \frac{1}{2} \left( \frac{1}{D^2 + 2D + 1} e^{0x} + \frac{1}{D^2 + 2D + 1} \cos 2x \right)
\]
Substitute \( D = 0 \) for the first term and \( D^2 = -4 \) for the second term:
\[
\frac{1}{D^2 + 2D + 1} e^{0x} = \frac{1}{1} = 1
\]
\[
\frac{1}{D^2 + 2D + 1} \cos 2x = \frac{1}{-4 + 2D + 1} \cos 2x = \frac{2D + 3}{4D^2 - 9} \cos 2x
\]
Substitute \( D^2 = -4 \):
\[
\frac{2D + 3}{4(-4) - 9} \cos 2x = \frac{2D + 3}{-25} \cos 2x
\]
Thus, the complete solution is:
\[
y = \text{C.F.} + \text{P.I.} = (c_1 + c_2 x) e^{-x} + \frac{1}{9} e^{2x} - \frac{1}{2} \left( 1 + \frac{2D + 3}{-25} \cos 2x \right)
\]
Simplify further:
\[
y = (c_1 + c_2 x) e^{-x} + \frac{1}{9} e^{2x} - \frac{1}{2} + \frac{3 \cos 2x - 4 \sin 2x}{50}
\]
3. Solve the differential equation: \( (D^2 - 4D + 3)y = \sin 3x \cos 2xx \)
Solution:
The given equation is in the form \( f(D)y = Q(x) \), where:
\[
f(D) = D^2 - 4D + 3 \quad \text{and} \quad Q(x) = \sin 3x \cos 2x
\]
The auxiliary equation is:
\[
f(m) = 0 \Rightarrow m^2 - 4m + 3 = 0 \Rightarrow m = 1, 3
\]
Thus, the complementary function (C.F.) is:
\[
\text{C.F.} = c_1 e^x + c_2 e^{3x}
\]
The particular integral (P.I.) is:
\[
\text{P.I.} = \frac{1}{D^2 - 4D + 3} \sin 3x \cos 2x
\]
Use the identity \( \sin 3x \cos 2x = \frac{1}{2} (\sin 5x + \sin x) \):
\[
\text{P.I.} = \frac{1}{2} \left( \frac{1}{D^2 - 4D + 3} \sin 5x + \frac{1}{D^2 - 4D + 3} \sin x \right)
\]
Substitute \( D^2 = -a^2 \) for each term:
\[
\frac{1}{D^2 - 4D + 3} \sin 5x = \frac{1}{-25 - 4D + 3} \sin 5x = \frac{1}{-22 - 4D} \sin 5x
\]
\[
\frac{1}{D^2 - 4D + 3} \sin x = \frac{1}{-1 - 4D + 3} \sin x = \frac{1}{2 - 4D} \sin x
\]
Simplify the expressions:
\[
\frac{1}{-22 - 4D} \sin 5x = \frac{4D - 22}{16D^2 - 484} \sin 5x = \frac{4D - 22}{-884} \sin 5x
\]
\[
\frac{1}{2 - 4D} \sin x = \frac{4D + 2}{16D^2 - 4} \sin x = \frac{4D + 2}{-20} \sin x
\]
Substitute \( D^2 = -a^2 \) again:
\[
\frac{4D - 22}{-884} \sin 5x = \frac{4(5 \cos 5x) - 22 \sin 5x}{-884} = \frac{20 \cos 5x - 22 \sin 5x}{-884}
\]
\[
\frac{4D + 2}{-20} \sin x = \frac{4 \cos x + 2 \sin x}{-20}
\]
Thus, the complete solution is:
\[
y = \text{C.F.} + \text{P.I.} = c_1 e^x + c_2 e^{3x} + \frac{10 \cos 5x - 11 \sin 5x}{884} + \frac{2 \cos x + 2 \sin x}{20}
\]
Model Problems
Solve the following differential equations:
-
Solve: \((D^2 - 4D + 1)y = \sin(2x + 3)\)
Answer: \(y = c_1 e^{(2 - \sqrt{3})x} + c_2 e^{(2 + \sqrt{3})x} + \frac{1}{73} \left[ 8\cos(2x + 3) - 3\sin(2x + 3) \right]\)
-
Solve: \((D^2 + 4)y = \sin^2 x\)
Answer: \(y = c_1 \cos 2x + c_2 \sin 2x + \frac{1}{8} - \frac{x}{8} \sin 2x\)
-
Solve: \(y'' + 4y' + 4y = 4\cos x + 3\sin x\) with \(y(0) = 1\), \(y'(0) = 0\)
Answer: \(y = (1 + x) e^{-2x} + \sin x\)
-
Solve: \((D^2 + 1)y = \sin x \sin 2x\)
Answer: \(y = (c_1 \cos x + c_2 \sin x) - \frac{x}{4} \sin x + \frac{1}{16} \cos 3x\)
-
Solve: \((D^3 - 2D^2 + 4D)y = e^{2x} + 2\sin 2x\)
Answer: \(y = e^x \left[ c_1 \cos \sqrt{3}x + c_3 \sin \sqrt{3}x \right] + c_3 + \frac{1}{8} e^{2x} + \frac{\sin 2x}{8}\)
-
Solve: \((D^3 - 1)y = e^x + \sin 3x + 2\)
Answer: \(y = c_1 e^x + e^{-\frac{x}{2}} \left[ c_2 \cos \left( \frac{\sqrt{3}}{2} x \right) + c_3 \sin \left( \frac{\sqrt{3}}{2} x \right) \right] + \frac{x}{3} e^x - 2 + \frac{1}{730} (27\cos 3x - \sin 3x)\)
-
Solve: \((D^2 - 8D + 9)y = 8\sin 5x\)
Answer: \(y = c_1 e^{(4 - \sqrt{7})x} + c_2 e^{(4 + \sqrt{7})x} + \frac{1}{29} \left[ 5\cos 5x - 2\sin 5x \right]\)
-
Solve: \((D^4 - k^4)y = \sin kx\)
Answer: \(y = c_1 e^{-kx} + c_2 e^{kx} + c_3 \cos kx + c_4 \sin kx + \frac{x}{4k^3} \cos kx\)
Consider the differential equation:
\[ f(D)y = Q(x) \]
where \( Q(x) = \sin(ax + b) \) or \( \cos(ax + b) \).
The particular integral (P.I.) is given by:
\[ \text{P.I.} = \frac{1}{f(D)} Q(x) = \frac{1}{f(D)} \sin(ax + b) \quad \text{or} \quad \frac{1}{f(D)} \cos(ax + b) \]
To evaluate this, substitute \( D^2 = -a^2 \):
\[ \text{P.I.} = \frac{1}{f(-a^2)} \sin(ax + b) \quad \text{or} \quad \frac{1}{f(-a^2)} \cos(ax + b) \]
This is valid provided \( f(-a^2) \neq 0 \).
If \( f(-a^2) = 0 \), then:
\[ \text{P.I.} = x \frac{1}{f'(D)} \sin(ax + b) \quad \text{or} \quad x \frac{1}{f'(D)} \cos(ax + b) \]
Substitute \( D^2 = -a^2 \):
\[ \text{P.I.} = x \frac{1}{f'(-a^2)} \sin(ax + b) \quad \text{or} \quad x \frac{1}{f'(-a^2)} \cos(ax + b) \]
This is valid provided \( f'(-a^2) \neq 0 \).
If \( f'(-a^2) = 0 \), then:
\[ \text{P.I.} = x^2 \frac{1}{f''(D)} \sin(ax + b) \quad \text{or} \quad x^2 \frac{1}{f''(D)} \cos(ax + b) \]
Substitute \( D^2 = -a^2 \):
\[ \text{P.I.} = x^2 \frac{1}{f''(-a^2)} \sin(ax + b) \quad \text{or} \quad x^2 \frac{1}{f''(-a^2)} \cos(ax + b) \]
This is valid provided \( f''(-a^2) \neq 0 \).
If \( f''(-a^2) = 0 \), continue the same process by increasing the power of \( x \) and differentiating \( f(D) \) further until a non-zero denominator is obtained.
1. Solve the differential equation: \( (D^4 + 10D^2 + 9)y = \cos(2x + 3) \)
Solution:
The given equation is in the form \( f(D)y = Q(x) \), where:
\[ f(D) = D^4 + 10D^2 + 9 \quad \text{and} \quad Q(x) = \cos(2x + 3) \]
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^4 + 10m^2 + 9 = 0 \Rightarrow m = \pm i, \pm 3i \]
Thus, the complementary function (C.F.) is:
\[ \text{C.F.} = (c_1 \cos x + c_2 \sin x) + (c_3 \cos 3x + c_4 \sin 3x) \]
The particular integral (P.I.) is:
\[ \text{P.I.} = \frac{1}{D^4 + 10D^2 + 9} \cos(2x + 3) \]
Substitute \( D^2 = -a^2 = -4 \) (since \( a = 2 \)):
\[ \text{P.I.} = \frac{1}{(-4)^2 + 10(-4) + 9} \cos(2x + 3) = \frac{1}{16 - 40 + 9} \cos(2x + 3) = -\frac{1}{15} \cos(2x + 3) \]
The complete solution is:
\[ y = \text{C.F.} + \text{P.I.} = (c_1 \cos x + c_2 \sin x) + (c_3 \cos 3x + c_4 \sin 3x) - \frac{1}{15} \cos(2x + 3) \]
2. Solve the differential equation: \( \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} + y = e^{2x} - \cos^2 x \)
Solution:
The given equation can be written as:
\[ (D^2 + 2D + 1)y = e^{2x} - \cos^2 x \]
This is in the form \( f(D)y = Q(x) \), where:
\[ f(D) = D^2 + 2D + 1 \quad \text{and} \quad Q(x) = e^{2x} - \cos^2 x \]
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^2 + 2m + 1 = 0 \Rightarrow m = -1, -1 \]
Thus, the complementary function (C.F.) is:
\[ \text{C.F.} = (c_1 + c_2 x) e^{-x} \]
The particular integral (P.I.) is:
\[ \text{P.I.} = \frac{1}{D^2 + 2D + 1} \left( e^{2x} - \cos^2 x \right) \]
Simplify \( \cos^2 x \) using the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \):
\[ \text{P.I.} = \frac{1}{D^2 + 2D + 1} e^{2x} - \frac{1}{D^2 + 2D + 1} \left( \frac{1 + \cos 2x}{2} \right) \]
Evaluate each term separately:
\[ \frac{1}{D^2 + 2D + 1} e^{2x} = \frac{1}{(2)^2 + 2(2) + 1} e^{2x} = \frac{1}{9} e^{2x} \]
\[ \frac{1}{D^2 + 2D + 1} \left( \frac{1 + \cos 2x}{2} \right) = \frac{1}{2} \left( \frac{1}{D^2 + 2D + 1} e^{0x} + \frac{1}{D^2 + 2D + 1} \cos 2x \right) \]
Substitute \( D = 0 \) for the first term and \( D^2 = -4 \) for the second term:
\[ \frac{1}{D^2 + 2D + 1} e^{0x} = \frac{1}{1} = 1 \]
\[ \frac{1}{D^2 + 2D + 1} \cos 2x = \frac{1}{-4 + 2D + 1} \cos 2x = \frac{2D + 3}{4D^2 - 9} \cos 2x \]
Substitute \( D^2 = -4 \):
\[ \frac{2D + 3}{4(-4) - 9} \cos 2x = \frac{2D + 3}{-25} \cos 2x \]
Thus, the complete solution is:
\[ y = \text{C.F.} + \text{P.I.} = (c_1 + c_2 x) e^{-x} + \frac{1}{9} e^{2x} - \frac{1}{2} \left( 1 + \frac{2D + 3}{-25} \cos 2x \right) \]
Simplify further:
\[ y = (c_1 + c_2 x) e^{-x} + \frac{1}{9} e^{2x} - \frac{1}{2} + \frac{3 \cos 2x - 4 \sin 2x}{50} \]
3. Solve the differential equation: \( (D^2 - 4D + 3)y = \sin 3x \cos 2xx \)
Solution:
The given equation is in the form \( f(D)y = Q(x) \), where:
\[ f(D) = D^2 - 4D + 3 \quad \text{and} \quad Q(x) = \sin 3x \cos 2x \]
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^2 - 4m + 3 = 0 \Rightarrow m = 1, 3 \]
Thus, the complementary function (C.F.) is:
\[ \text{C.F.} = c_1 e^x + c_2 e^{3x} \]
The particular integral (P.I.) is:
\[ \text{P.I.} = \frac{1}{D^2 - 4D + 3} \sin 3x \cos 2x \]
Use the identity \( \sin 3x \cos 2x = \frac{1}{2} (\sin 5x + \sin x) \):
\[ \text{P.I.} = \frac{1}{2} \left( \frac{1}{D^2 - 4D + 3} \sin 5x + \frac{1}{D^2 - 4D + 3} \sin x \right) \]
Substitute \( D^2 = -a^2 \) for each term:
\[ \frac{1}{D^2 - 4D + 3} \sin 5x = \frac{1}{-25 - 4D + 3} \sin 5x = \frac{1}{-22 - 4D} \sin 5x \]
\[ \frac{1}{D^2 - 4D + 3} \sin x = \frac{1}{-1 - 4D + 3} \sin x = \frac{1}{2 - 4D} \sin x \]
Simplify the expressions:
\[ \frac{1}{-22 - 4D} \sin 5x = \frac{4D - 22}{16D^2 - 484} \sin 5x = \frac{4D - 22}{-884} \sin 5x \]
\[ \frac{1}{2 - 4D} \sin x = \frac{4D + 2}{16D^2 - 4} \sin x = \frac{4D + 2}{-20} \sin x \]
Substitute \( D^2 = -a^2 \) again:
\[ \frac{4D - 22}{-884} \sin 5x = \frac{4(5 \cos 5x) - 22 \sin 5x}{-884} = \frac{20 \cos 5x - 22 \sin 5x}{-884} \]
\[ \frac{4D + 2}{-20} \sin x = \frac{4 \cos x + 2 \sin x}{-20} \]
Thus, the complete solution is:
\[ y = \text{C.F.} + \text{P.I.} = c_1 e^x + c_2 e^{3x} + \frac{10 \cos 5x - 11 \sin 5x}{884} + \frac{2 \cos x + 2 \sin x}{20} \]
Model Problems
Solve the following differential equations:
-
Solve: \((D^2 - 4D + 1)y = \sin(2x + 3)\)
Answer: \(y = c_1 e^{(2 - \sqrt{3})x} + c_2 e^{(2 + \sqrt{3})x} + \frac{1}{73} \left[ 8\cos(2x + 3) - 3\sin(2x + 3) \right]\) -
Solve: \((D^2 + 4)y = \sin^2 x\)
Answer: \(y = c_1 \cos 2x + c_2 \sin 2x + \frac{1}{8} - \frac{x}{8} \sin 2x\) -
Solve: \(y'' + 4y' + 4y = 4\cos x + 3\sin x\) with \(y(0) = 1\), \(y'(0) = 0\)
Answer: \(y = (1 + x) e^{-2x} + \sin x\) -
Solve: \((D^2 + 1)y = \sin x \sin 2x\)
Answer: \(y = (c_1 \cos x + c_2 \sin x) - \frac{x}{4} \sin x + \frac{1}{16} \cos 3x\) -
Solve: \((D^3 - 2D^2 + 4D)y = e^{2x} + 2\sin 2x\)
Answer: \(y = e^x \left[ c_1 \cos \sqrt{3}x + c_3 \sin \sqrt{3}x \right] + c_3 + \frac{1}{8} e^{2x} + \frac{\sin 2x}{8}\) -
Solve: \((D^3 - 1)y = e^x + \sin 3x + 2\)
Answer: \(y = c_1 e^x + e^{-\frac{x}{2}} \left[ c_2 \cos \left( \frac{\sqrt{3}}{2} x \right) + c_3 \sin \left( \frac{\sqrt{3}}{2} x \right) \right] + \frac{x}{3} e^x - 2 + \frac{1}{730} (27\cos 3x - \sin 3x)\) -
Solve: \((D^2 - 8D + 9)y = 8\sin 5x\)
Answer: \(y = c_1 e^{(4 - \sqrt{7})x} + c_2 e^{(4 + \sqrt{7})x} + \frac{1}{29} \left[ 5\cos 5x - 2\sin 5x \right]\) -
Solve: \((D^4 - k^4)y = \sin kx\)
Answer: \(y = c_1 e^{-kx} + c_2 e^{kx} + c_3 \cos kx + c_4 \sin kx + \frac{x}{4k^3} \cos kx\)