Case I: When \( Q(x) = e^{ax} \)
If \( Q(x) = e^{ax} \) (where \( a \) is real or complex), the Particular Integral (P.I.) is given by:
\[ \text{P.I.} = \frac{1}{f(D)} Q(x) = \frac{1}{f(D)} e^{ax} \]
To evaluate this, substitute \( D = a \):
\[ \text{P.I.} = \frac{1}{f(a)} e^{ax} \quad \text{(provided that } f(a) \neq 0\text{)} \]
If \( f(a) = 0 \), then:
\[ \text{P.I.} = x \frac{1}{f'(D)} e^{ax} \]
Substitute \( D = a \):
\[ \text{P.I.} = x \frac{1}{f'(a)} e^{ax} \quad \text{(provided that } f'(a) \neq 0\text{)} \]
If \( f'(a) = 0 \), then:
\[ \text{P.I.} = x^2 \frac{1}{f''(D)} e^{ax} \]
Substitute \( D = a \):
\[ \text{P.I.} = x^2 \frac{1}{f''(a)} e^{ax} \quad \text{(provided that } f''(a) \neq 0\text{)} \]
If \( f''(a) = 0 \), 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: \( (D^2 - 5D + 6)y = e^{4x} \)
Solution:
The given equation is in the form \( f(D)y = Q(x) \), where:
\[ f(D) = D^2 - 5D + 6 \quad \text{and} \quad Q(x) = e^{4x} \]
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^2 - 5m + 6 = 0 \Rightarrow m = 2, 3 \]
Thus, the complementary function (C.F.) is:
\[ \text{C.F.} = c_1 e^{2x} + c_2 e^{3x} \]
The particular integral (P.I.) is:
\[ \text{P.I.} = \frac{1}{f(D)} Q(x) = \frac{1}{D^2 - 5D + 6} e^{4x} \quad \text{(put } D = 4\text{)} \]
\[ = \frac{1}{16 - 20 + 6} e^{4x} = \frac{1}{2} e^{4x} \]
The complete solution is:
\[ y = \text{C.F.} + \text{P.I.} = c_1 e^{2x} + c_2 e^{3x} + \frac{1}{2} e^{4x} \]
2. Solve the differential equation: \( (D^3 + 6D^2 + 9D)y = e^{-3x} \)
Solution:
The given equation is in the form \( f(D)y = Q(x) \), where:
\[ f(D) = D^3 + 6D^2 + 9D \quad \text{and} \quad Q(x) = e^{-3x} \]
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^3 + 6m^2 + 9m = 0 \Rightarrow m = -3, -3, 0 \]
Thus, the complementary function (C.F.) is:
\[ \text{C.F.} = (c_1 + c_2 x) e^{-3x} + c_3 \]
The particular integral (P.I.) is:
\[ \text{P.I.} = \frac{1}{f(D)} Q(x) = \frac{1}{D^3 + 6D^2 + 9D} e^{-3x} \quad \text{(put } D = -3\text{)} \]
Since \( f(-3) = 0 \), we proceed as follows:
\[ \text{P.I.} = x \frac{1}{f'(D)} e^{-3x} = x \frac{1}{3D^2 + 12D + 9} e^{-3x} \quad \text{(put } D = -3\text{)} \]
Since \( f'(-3) = 0 \), we proceed further:
\[ \text{P.I.} = x^2 \frac{1}{f''(D)} e^{-3x} = x^2 \frac{1}{6D + 12} e^{-3x} \quad \text{(put } D = -3\text{)} \]
\[ = -\frac{x^2 e^{-3x}}{6} \]
The complete solution is:
\[ y = \text{C.F.} + \text{P.I.} = (c_1 + c_2 x) e^{-3x} + c_3 - \frac{x^2 e^{-3x}}{6} \]
3. Solve the differential equation: \( \frac{d^2 y}{dx^2} - 6 \frac{dy}{dx} + 9y = 6e^{3x} + 7e^{-2x} - \log 2 \)
Solution:
The given equation can be written as:
\[ (D^2 - 6D + 9)y = 6e^{3x} + 7e^{-2x} - \log 2 \]
This is in the form \( f(D)y = Q(x) \), where:
\[ f(D) = D^2 - 6D + 9 \quad \text{and} \quad Q(x) = 6e^{3x} + 7e^{-2x} - \log 2 \]
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^2 - 6m + 9 = 0 \Rightarrow m = 3, 3 \]
Thus, the complementary function (C.F.) is:
\[ \text{C.F.} = (c_1 + c_2 x) e^{3x} \]
The particular integral (P.I.) is:
\[ \text{P.I.} = \frac{1}{D^2 - 6D + 9} \left( 6e^{3x} + 7e^{-2x} - \log 2 \right) \]
\[ = 6 \frac{1}{D^2 - 6D + 9} e^{3x} + 7 \frac{1}{D^2 - 6D + 9} e^{-2x} - \log 2 \frac{1}{D^2 - 6D + 9} e^{0x} \]
Evaluating each term separately:
\[ 6 \frac{1}{D^2 - 6D + 9} e^{3x} = 6x \frac{1}{f'(D)} e^{3x} = 6x^2 \frac{1}{f''(D)} e^{3x} = 3x^2 e^{3x} \]
\[ 7 \frac{1}{D^2 - 6D + 9} e^{-2x} = \frac{7}{25} e^{-2x} \]
\[ -\log 2 \frac{1}{D^2 - 6D + 9} e^{0x} = -\frac{\log 2}{9} \]
Thus, the complete solution is:
\[ y = \text{C.F.} + \text{P.I.} = (c_1 + c_2 x) e^{3x} + 3x^2 e^{3x} + \frac{7}{25} e^{-2x} - \frac{\log 2}{9} \]
4. Solve the differential equation: \( (D^2 + 6D + 9)y = 5^x - \log 2 \)
Solution:
The given equation is in the form \( f(D)y = Q(x) \), where:
\[ f(D) = D^2 + 6D + 9 \quad \text{and} \quad Q(x) = 5^x - \log 2 \]
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^2 + 6m + 9 = 0 \Rightarrow m = -3, -3 \]
Thus, the complementary function (C.F.) is:
\[ \text{C.F.} = (c_1 + c_2 x) e^{-3x} \]
The particular integral (P.I.) is:
\[ \text{P.I.} = \frac{1}{D^2 + 6D + 9} (5^x - \log 2) \]
\[ = \frac{1}{D^2 + 6D + 9} 5^x - \log 2 \frac{1}{D^2 + 6D + 9} e^{0x} \]
Evaluating each term separately:
\[ \frac{1}{D^2 + 6D + 9} 5^x = \frac{1}{(\log 5)^2 + 6 \log 5 + 9} 5^x \]
\[ -\log 2 \frac{1}{D^2 + 6D + 9} e^{0x} = -\frac{\log 2}{9} \]
Thus, the complete solution is:
\[ y = \text{C.F.} + \text{P.I.} = (c_1 + c_2 x) e^{-3x} + \frac{1}{(\log 5)^2 + 6 \log 5 + 9} 5^x - \frac{\log 2}{9} \]
5. Solve the differential equation: \( \frac{d^2 y}{dx^2} + 4 \frac{dy}{dx} + 5y = -2 \cosh x \)
Solution:
The given equation can be written as:
\[ (D^2 + 4D + 5)y = -2 \cosh x \]
This is in the form \( f(D)y = Q(x) \), where:
\[ f(D) = D^2 + 4D + 5 \quad \text{and} \quad Q(x) = -2 \cosh x \]
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^2 + 4m + 5 = 0 \Rightarrow m = -2 \pm i \]
Thus, the complementary function (C.F.) is:
\[ \text{C.F.} = e^{-2x} (c_1 \cos x + c_2 \sin x) \]
The particular integral (P.I.) is:
\[ \text{P.I.} = \frac{1}{D^2 + 4D + 5} (-2 \cosh x) \]
\[ = -\frac{2}{D^2 + 4D + 5} \left( \frac{e^x + e^{-x}}{2} \right) \]
\[ = -\frac{1}{D^2 + 4D + 5} e^x - \frac{1}{D^2 + 4D + 5} e^{-x} \]
Evaluating each term separately:
\[ -\frac{1}{D^2 + 4D + 5} e^x = -\frac{1}{10} e^x \]
\[ -\frac{1}{D^2 + 4D + 5} e^{-x} = -\frac{1}{2} e^{-x} \]
Thus, the complete solution is:
\[ y = \text{C.F.} + \text{P.I.} = e^{-2x} (c_1 \cos x + c_2 \sin x) - \frac{1}{10} e^x - \frac{1}{2} e^{-x} \]
Model Problems
Solve the following differential equations:
-
Solve: \((D^3 + 3D^2 + 3D + 1)y = e^{-x}\)
Answer: \(y = (c_1 + c_2 x + c_3 x^3) e^{-x} + \frac{x^3}{6} e^{-x}\) -
Solve: \(\frac{d^3y}{dx^3} - y = (e^x + 1)^2\)
Answer: \(y = c_1 e^x + e^{-\frac{x}{2}} \left[ c_2 \cos \left( \frac{\sqrt{3}}{2} x \right) + c_3 \sin \left( \frac{\sqrt{3}}{2} x \right) \right] + \frac{1}{7} e^{2x} + \frac{2x}{3} - 1\) -
Solve: \((D^3 + 1)y = 3 + 5e^x\)
Answer: \(y = c_1 e^{-x} + e^{\frac{x}{2}} \left[ c_2 \cos \left( \frac{\sqrt{3}}{2} x \right) + c_3 \sin \left( \frac{\sqrt{3}}{2} x \right) \right] + 3 + \frac{5}{2} e^x\) -
Solve: \(\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 5y = -2\cosh x\) with \(y(0) = 0\), \(y'(0) = 1\)
Answer: \(y = \frac{3}{5} e^{-2x} (\cos x + 2\sin x) - \frac{e^x}{10} - \frac{e^{-x}}{2}\) -
Solve: \((D + 2)(D - 1)^2 y = e^{-2x} + 2\sinh x\)
Answer: \(y = c_1 e^{-2x} + (c_2 + c_3 x) e^x + \frac{x}{9} e^{-2x} - \frac{x^2}{6} e^x - \frac{1}{4} e^{-x}\) -
Solve: \(\frac{d^2y}{dx^2} - p^2 y = \sinh px\)
Answer: \(y = c_1 e^{-px} + c_2 e^{px} + \frac{x}{2p} \cosh px\) -
Solve: \((D^3 - 12D + 16)y = (e^x + e^{-2x})^2\)
Answer: \(y = (c_1 + c_2 x) e^{2x} + c_3 e^{-4x} + \frac{x^2}{12} e^{2x} - \frac{2}{27} e^{-x} + \frac{x}{36} e^{-4x}\) -
Solve: \(D(D + 1)^2 y = 12e^{-x}\)
Answer: \(y = c_1 + (c_2 + c_3 x) e^{-x} - 6x^2 e^{-x}\)