Case III: Particular Integral for Polynomial Functions When \( Q(x) = x^m \) (where \( m \) is a positive integer)
Consider the differential equation:
\[ f(D)y = Q(x) \]
where \( Q(x) = x^m \) and \( m \) is a positive integer.
The particular integral (P.I.) is given by:
\[ \text{P.I.} = \frac{1}{f(D)} Q(x) = \frac{1}{f(D)} x^m \]
To evaluate this, write \( f(D) \) in the form of \( \frac{1}{[1 \pm \phi(D)]} \) or \( [1 \pm \phi(D)]^{-1} \) by taking out the lowest degree term from \( f(D) \). Then, expand it in ascending powers of \( D \) up to the term containing \( D^m \), and operate on \( x^m \) term by term. (Note that \( D^{m+1} x^m = 0 \)).
Thus:
\[ \text{P.I.} = \frac{1}{[1 \pm \phi(D)]} x^m = [1 \pm \phi(D)]^{-1} x^m \]
Important Expansions:
1) \( (1 + x)^{-1} = 1 - x + x^2 - x^3 + x^4 - \dots \)
2) \( (1 - x)^{-1} = 1 + x + x^2 + x^3 + x^4 + \dots \)
3) \( (1 + x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + 5x^4 - \dots \)
4) \( (1 - x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + \dots \)
1. Solve the differential equation: \( (D^2 + D)y = x^2 + 2x + 4 \)
Solution:
The given equation is in the form \( f(D)y = Q(x) \), where:
\[ f(D) = D^2 + D \quad \text{and} \quad Q(x) = x^2 + 2x + 4 \]
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^2 + m = 0 \Rightarrow m = 0, -1 \]
Thus, the complementary function (C.F.) is:
\[ \text{C.F.} = c_1 e^{0x} + c_2 e^{-x} = c_1 + c_2 e^{-x} \]
The particular integral (P.I.) is:
\[ \text{P.I.} = \frac{1}{D^2 + D} (x^2 + 2x + 4) \]
Rewrite \( \frac{1}{D^2 + D} \) as \( \frac{1}{D(1 + D)} \):
\[ \text{P.I.} = \frac{1}{D} \cdot \frac{1}{1 + D} (x^2 + 2x + 4) \]
Expand \( \frac{1}{1 + D} \) using the binomial expansion:
\[ \frac{1}{1 + D} = 1 - D + D^2 - D^3 + \dots \]
Apply the expansion to \( x^2 + 2x + 4 \):
\[ \text{P.I.} = \frac{1}{D} \left[ (x^2 + 2x + 4) - D(x^2 + 2x + 4) + D^2(x^2 + 2x + 4) \right] \]
Compute the derivatives:
\[ D(x^2 + 2x + 4) = 2x + 2 \]
\[ D^2(x^2 + 2x + 4) = 2 \]
Substitute back:
\[ \text{P.I.} = \frac{1}{D} \left[ (x^2 + 2x + 4) - (2x + 2) + 2 \right] = \frac{1}{D} (x^2 + 4) \]
Integrate \( x^2 + 4 \) with respect to \( x \):
\[ \text{P.I.} = \int (x^2 + 4) \, dx = \frac{x^3}{3} + 4x \]
The complete solution is:
\[ y = \text{C.F.} + \text{P.I.} = c_1 + c_2 e^{-x} + \frac{x^3}{3} + 4x \]
2. Solve the differential equation: \( (D^3 - 2D + 4)y = x^4 + 3x^2 - 5x + 2 \)
Solution:
The given equation is in the form \( f(D)y = Q(x) \), where:
\[ f(D) = D^3 - 2D + 4 \quad \text{and} \quad Q(x) = x^4 + 3x^2 - 5x + 2 \]
The auxiliary equation is:
\[ f(m) = 0 \Rightarrow m^3 - 2m + 4 = 0 \Rightarrow m = 1 \pm i, -2 \]
Thus, the complementary function (C.F.) is:
\[ \text{C.F.} = e^x (c_1 \cos x + c_2 \sin x) + c_3 e^{-2x} \]
The particular integral (P.I.) is:
\[ \text{P.I.} = \frac{1}{D^3 - 2D + 4} (x^4 + 3x^2 - 5x + 2) \]
Rewrite \( \frac{1}{D^3 - 2D + 4} \) as \( \frac{1}{4(1 + \frac{D^3 - 2D}{4})} \):
\[ \text{P.I.} = \frac{1}{4} \left( 1 + \frac{D^3 - 2D}{4} \right)^{-1} (x^4 + 3x^2 - 5x + 2) \]
Expand \( \left( 1 + \frac{D^3 - 2D}{4} \right)^{-1} \) using the binomial expansion:
\[ \left( 1 + \frac{D^3 - 2D}{4} \right)^{-1} = 1 - \frac{D^3 - 2D}{4} + \left( \frac{D^3 - 2D}{4} \right)^2 - \left( \frac{D^3 - 2D}{4} \right)^3 + \dots \]
Apply the expansion to \( x^4 + 3x^2 - 5x + 2 \):
\[ \text{P.I.} = \frac{1}{4} \left[ 1 - \frac{D^3 - 2D}{4} + \left( \frac{D^3 - 2D}{4} \right)^2 - \left( \frac{D^3 - 2D}{4} \right)^3 + \dots \right] (x^4 + 3x^2 - 5x + 2) \]
Compute the derivatives up to \( D^4 \) (since higher derivatives of \( x^4 \) will vanish):
\[ D(x^4 + 3x^2 - 5x + 2) = 4x^3 + 6x - 5 \]
\[ D^2(x^4 + 3x^2 - 5x + 2) = 12x^2 + 6 \]
\[ D^3(x^4 + 3x^2 - 5x + 2) = 24x \]
\[ D^4(x^4 + 3x^2 - 5x + 2) = 24 \]
Substitute back:
\[ \text{P.I.} = \frac{1}{4} \left[ (x^4 + 3x^2 - 5x + 2) - \frac{24x - 2(4x^3 + 6x - 5)}{4} + \frac{(12x^2 + 6)}{4} - \frac{24}{8} \right] \]
Simplify the expression:
\[ \text{P.I.} = \frac{1}{4} \left[ x^4 + 3x^2 - 5x + 2 + \frac{4x^3 + 6x - 5}{2} + \frac{12x^2 + 6}{4} - 3 \right] \]
Combine like terms:
\[ \text{P.I.} = \frac{1}{4} \left[ x^4 + 2x^3 + 6x^2 - 5x - \frac{7}{2} \right] = \frac{1}{8} (2x^4 + 4x^3 + 12x^2 - 10x - 7) \]
The complete solution is:
\[ y = \text{C.F.} + \text{P.I.} = e^x (c_1 \cos x + c_2 \sin x) + c_3 e^{-2x} + \frac{1}{8} (2x^4 + 4x^3 + 12x^2 - 10x - 7) \]
3. Solve the differential equation: \( (D^2 - 4D + 4)y = 8(e^{2x} + \sin 2x + x^2) \)
Solution:
The auxiliary equation is:
\[ m^2 - 4m + 4 = 0 \Rightarrow m = 2, 2 \]
Thus, the complementary function (C.F.) is:
\[ \text{C.F.} = (c_1 + c_2 x) e^{2x} \]
The particular integral (P.I.) is:
\[ \text{P.I.} = \frac{8}{D^2 - 4D + 4} (e^{2x} + \sin 2x + x^2) \]
We can split the P.I. into three parts:
\[ \text{P.I.} = 8 \left( \text{P.I.}_1 + \text{P.I.}_2 + \text{P.I.}_3 \right) \]
Where:
\[ \text{P.I.}_1 = \frac{1}{D^2 - 4D + 4} e^{2x} \]
\[ \text{P.I.}_2 = \frac{1}{D^2 - 4D + 4} \sin 2x \]
\[ \text{P.I.}_3 = \frac{1}{D^2 - 4D + 4} x^2 \]
1. Calculation of \( \text{P.I.}_1 \):
For \( \text{P.I.}_1 \):
\[ \text{P.I.}_1 = \frac{1}{(D - 2)^2} e^{2x} \]
Since \( D = 2 \) is a repeated root, we use:
\[ \text{P.I.}_1 = \frac{x^2}{2} e^{2x} \]
2. Calculation of \( \text{P.I.}_2 \):
For \( \text{P.I.}_2 \):
\[ \text{P.I.}_2 = \frac{1}{D^2 - 4D + 4} \sin 2x \]
Substitute \( D^2 = -4 \) (since \( a = 2 \)):
\[ \text{P.I.}_2 = \frac{1}{-4 - 4D + 4} \sin 2x = \frac{1}{-4D} \sin 2x \]
Integrate \( \sin 2x \):
\[ \text{P.I.}_2 = -\frac{1}{8} \cos 2x \]
3. Calculation of \( \text{P.I.}_3 \):
For \( \text{P.I.}_3 \):
\[ \text{P.I.}_3 = \frac{1}{D^2 - 4D + 4} x^2 \]
Rewrite \( \frac{1}{(D - 2)^2} \) as \( \frac{1}{4(1 - \frac{D}{2})^2} \):
\[ \text{P.I.}_3 = \frac{1}{4} \left( 1 - \frac{D}{2} \right)^{-2} x^2 \]
Expand \( \left( 1 - \frac{D}{2} \right)^{-2} \) using the binomial expansion:
\[ \left( 1 - \frac{D}{2} \right)^{-2} = 1 + D + \frac{3D^2}{4} + \dots \]
Apply the expansion to \( x^2 \):
\[ \text{P.I.}_3 = \frac{1}{4} \left( x^2 + D(x^2) + \frac{3D^2(x^2)}{4} \right) \]
Compute the derivatives:
\[ D(x^2) = 2x \]
\[ D^2(x^2) = 2 \]
Substitute back:
\[ \text{P.I.}_3 = \frac{1}{4} \left( x^2 + 2x + \frac{3}{2} \right) \]
Combine all parts of the particular integral:
\[ \text{P.I.} = 8 \left( \frac{x^2}{2} e^{2x} - \frac{1}{8} \cos 2x + \frac{1}{4} \left( x^2 + 2x + \frac{3}{2} \right) \right) \]
Simplify:
\[ \text{P.I.} = 4x^2 e^{2x} - \cos 2x + 2x^2 + 4x + 3 \]
The complete solution is:
\[ y = \text{C.F.} + \text{P.I.} = (c_1 + c_2 x) e^{2x} + 4x^2 e^{2x} - \cos 2x + 2x^2 + 4x + 3 \]
4. Solve the differential equation: \( y = \text{C.F.} + \text{P.I.} \)
Where:
\[ \text{C.F.} = e^x (c_1 \cos x + c_2 \sin x) + c_3 e^{-2x} \]
And:
\[ \text{P.I.} = \frac{x^2}{2} e^{2x} - \frac{1}{8} \cos 2x + \frac{1}{4} \left( x^2 + 2x + \frac{3}{2} \right) \]
Thus, the complete solution is:
\[ y = e^x (c_1 \cos x + c_2 \sin x) + c_3 e^{-2x} + \frac{x^2}{2} e^{2x} - \frac{1}{8} \cos 2x + \frac{1}{4} \left( x^2 + 2x + \frac{3}{2} \right) \]
Model Problems
Solve the following differential equations:
-
Solve: \((D^2 + 3D + 2)y = x^3 + x^2\)
Answer: \(y = c_1 e^{-x} + c_2 e^{-2x} + \frac{1}{8} \left[ 4x^3 - 14x^2 + 30x - 31 \right]\) -
Solve: \((D^4 + D^3 + D^2)y = 5x^2\)
Answer: \(y = c_1 + c_2 x + e^{-\frac{x}{2}} \left[ c_3 \cos \left( \frac{\sqrt{3}}{2} x \right) + c_4 \sin \left( \frac{\sqrt{3}}{2} x \right) \right] + \frac{5x^3}{12} (x - 4)\) -
Solve: \((D^3 - D)y = 2x + 1 + 4\cos x + 2e^x\)
Answer: \(y = c_1 e^{-x} + c_2 + c_3 e^x - x^2 - x - 2\sin x + x e^x\) -
Solve: \((D^2 + 1)^2 y = x^4 + 2\sin x \cos 3x\)
Answer: \(y = (c_1 + c_2 x) \cos x + (c_3 + c_4 x) \sin x + x^4 - 24x^2 - \frac{\sin 2x}{9} + \frac{\sin 4x}{225} + 72\) -
Solve: \((D^2 + D - 2)y = 2(1 + x - x^2)\)
Answer: \(y = c_1 e^x + c_2 e^{-2x} + x^2\)