Case V: Particular Integral When \( Q(x) = x V(x) \)
Consider the differential equation:
\[
f(D)y = Q(x)
\]
where \( Q(x) = x V(x) \), and \( V(x) = \sin(ax + b) \) or \( \cos(ax + b) \).
The particular integral (P.I.) is given by:
\[
\text{P.I.} = \frac{1}{f(D)} Q(x) = \frac{1}{f(D)} x V(x)
\]
Using the rule for \( Q(x) = x V(x) \):
\[
\text{P.I.} = \left[ x - \frac{1}{f(D)} f'(D) \right] \frac{1}{f(D)} V(x)
\]
Here, \( f'(D) \) is the derivative of \( f(D) \) with respect to \( D \).
Note:
For finding the particular integral of \( f(D)y = x^m \sin ax \) or \( x^m \cos ax \), we use the fact that:
\[
e^{iax} = \cos ax + i \sin ax
\]
Thus:
\[
\cos ax = \text{Real part of } e^{iax} = \text{R.P.}(e^{iax})
\]
\[
\sin ax = \text{Imaginary part of } e^{iax} = \text{I.P.}(e^{iax})
\]
1. Solve the differential equation: \( (D^2 + 2D + 1)y = x \cos x \)
Solution:
The given equation is in the form \( f(D)y = Q(x) \), where:
\[
f(D) = D^2 + 2D + 1 \quad \text{and} \quad Q(x) = x \cos x
\]
The auxiliary equation is:
\[
m^2 + 2m + 1 = 0 \Rightarrow m = -1, -1
\]
Thus, the complementary function (C.F.) is:
\[
\text{C.F.} = (c_1 + c_2 x) e^{-x}
\]
The particular integral (P.I.) is:
\[
\text{P.I.} = \frac{1}{D^2 + 2D + 1} x \cos x
\]
Using the rule for \( Q(x) = x V(x) \), where \( V(x) = \cos x \):
\[
\text{P.I.} = \left[ x - \frac{1}{f(D)} f'(D) \right] \frac{1}{f(D)} V(x)
\]
Here, \( f(D) = D^2 + 2D + 1 \) and \( f'(D) = 2D + 2 \).
Thus:
\[
\text{P.I.} = \left[ x - \frac{1}{D^2 + 2D + 1} (2D + 2) \right] \frac{1}{D^2 + 2D + 1} \cos x
\]
Simplify \( \frac{1}{D^2 + 2D + 1} \cos x \):
\[
\frac{1}{D^2 + 2D + 1} \cos x = \frac{1}{-1 + 2D + 1} \cos x = \frac{1}{2D} \cos x
\]
Integrate \( \cos x \):
\[
\frac{1}{2D} \cos x = \frac{1}{2} \int \cos x \, dx = \frac{1}{2} \sin x
\]
Now, compute \( \frac{1}{D^2 + 2D + 1} (2D + 2) \sin x \):
\[
\frac{1}{D^2 + 2D + 1} (2D + 2) \sin x = \frac{1}{D^2 + 2D + 1} (2 \cos x + 2 \sin x)
\]
Simplify further:
\[
\text{P.I.} = \left[ x - \frac{2 \cos x + 2 \sin x}{D^2 + 2D + 1} \right] \frac{1}{2} \sin x
\]
Evaluate \( \frac{2 \cos x + 2 \sin x}{D^2 + 2D + 1} \):
\[
\frac{2 \cos x + 2 \sin x}{D^2 + 2D + 1} = \frac{2 \cos x + 2 \sin x}{-1 + 2D + 1} = \frac{2 \cos x + 2 \sin x}{2D}
\]
Integrate:
\[
\frac{2 \cos x + 2 \sin x}{2D} = \frac{1}{D} (\cos x + \sin x) = \sin x - \cos x
\]
Thus:
\[
\text{P.I.} = \left[ x - (\sin x - \cos x) \right] \frac{1}{2} \sin x = \frac{x}{2} \sin x - \frac{1}{2} (\sin x - \cos x) \sin x
\]
Simplify the expression:
\[
\text{P.I.} = \frac{x}{2} \sin x - \frac{1}{2} \sin^2 x + \frac{1}{2} \cos x \sin x
\]
The complete solution is:
\[
y = \text{C.F.} + \text{P.I.} = (c_1 + c_2 x) e^{-x} + \frac{x}{2} \sin x + \frac{1}{2} (\cos x - \sin x)
\]
2. Solve the Differential Equation \(\left( {{D^2} + 1} \right)y = {x^2}\sin 2x\)
Solution
The auxiliary equation is:
\[
m^2 + 1 = 0 \implies m = \pm i
\]
Therefore, the complementary function (C.F.) is:
\[
\text{C.F.} = c_1 \cos x + c_2 \sin x
\]
Find the Particular Integral (P.I.)
To find the particular integral, we use the method of undetermined coefficients. The right-hand side of the equation is \(x^2 \sin 2x\), which can be expressed as the imaginary part of \(e^{2ix} x^2\):
\[
\text{P.I.} = \text{Imaginary Part of } e^{2ix} \frac{1}{(D + 2i)^2 + 1} x^2
\]
Replace \(D\) by \(D + 2i\):
\[
\text{P.I.} = \text{Imaginary Part of } e^{2ix} \frac{1}{-3\left[1 - \left(\frac{4iD + D^2}{3}\right)\right]} x^2
\]
Expand the denominator using the binomial series:
\[
\text{P.I.} = \text{Imaginary Part of } \frac{e^{2ix}}{-3} \left[1 + \left(\frac{4iD + D^2}{3}\right) + \left(\frac{4iD + D^2}{3}\right)^2 + \cdots \right] x^2
\]
Simplify the expansion:
\[
\text{P.I.} = \text{Imaginary Part of } \frac{e^{2ix}}{-3} \left[1 + \frac{4i}{3}D - \frac{13}{9}D^2 + \cdots \right] x^2
\]
Apply the differential operators:
\[
\text{P.I.} = \text{Imaginary Part of } \frac{e^{2ix}}{-3} \left[x^2 + \frac{4i}{3}(2x) - \frac{13}{9}(2)\right]
\]
Simplify further:
\[
\text{P.I.} = \text{Imaginary Part of } \left\{ \frac{-1}{3} (\cos 2x + i \sin 2x) \left[\left(x^2 - \frac{26}{9}\right) + i \frac{8x}{3}\right] \right\}
\]
Extract the imaginary part:
\[
\text{P.I.} = \frac{-1}{3} \left[\left(x^2 - \frac{26}{9}\right) \sin 2x + \frac{8}{3} x \cos 2x \right]
\]
The Complete Solution
The complete solution is the sum of the complementary function and the particular integral:
\[
y = c_1 \cos x + c_2 \sin x - \frac{1}{3} \left[\left(x^2 - \frac{26}{9}\right) \sin 2x + \frac{8}{3} x \cos 2x \right]
\]
Model Problems
Solve the following differential equations:
-
Solve: \(\frac{d^2y}{dx^2} + 16y = x \sin 3x\)
Answer: \(y = c_1 \cos 4x + c_2 \sin 4x + \frac{1}{49} (7x \sin 3x - 6 \cos 3x)\)
-
Solve: \(\frac{d^2y}{dx^2} - y = x \sin x + x^2 e^x\)
Answer: \(y = c_1 e^x + c_2 e^{-x} - \frac{1}{2} \left[ x \sin x + \cos x \right] + \frac{x e^x}{12} \left( 2x^2 - 3x + 3 \right)\)
-
Solve: \(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = x e^x \sin x\)
Answer: \(y = (c_1 + c_2 x) e^{-x} - e^x \left( x \sin x + 2 \cos x \right)\)
-
Solve: \((D^2 + 1)y = x \cos x\)
Answer: \(y = c_1 \cos x + c_2 \sin x + \frac{1}{4} \left( x^2 \sin x + x \cos x \right)\)
-
Solve: \((D^2 - 1)y = x^2 \cos x\)
Answer: \(y = c_1 e^{-x} + c_2 e^x - \frac{1}{2} \left( x^2 \cos x - 2x \sin x \right)\)
-
Solve: \((D^2 - 4D + 4)y = 8x^2 e^{2x} \sin 2x\)
Answer: \(y = (c_1 + c_2 x) e^{2x} + e^{2x} \left[ (3 - 2x^2) \sin 2x - 4x \cos 2x \right]\)
Consider the differential equation:
\[ f(D)y = Q(x) \]
where \( Q(x) = x V(x) \), and \( V(x) = \sin(ax + b) \) or \( \cos(ax + b) \).
The particular integral (P.I.) is given by:
\[ \text{P.I.} = \frac{1}{f(D)} Q(x) = \frac{1}{f(D)} x V(x) \]
Using the rule for \( Q(x) = x V(x) \):
\[ \text{P.I.} = \left[ x - \frac{1}{f(D)} f'(D) \right] \frac{1}{f(D)} V(x) \]
Here, \( f'(D) \) is the derivative of \( f(D) \) with respect to \( D \).
Note:
For finding the particular integral of \( f(D)y = x^m \sin ax \) or \( x^m \cos ax \), we use the fact that:
\[ e^{iax} = \cos ax + i \sin ax \]
Thus:
\[ \cos ax = \text{Real part of } e^{iax} = \text{R.P.}(e^{iax}) \]
\[ \sin ax = \text{Imaginary part of } e^{iax} = \text{I.P.}(e^{iax}) \]
1. Solve the differential equation: \( (D^2 + 2D + 1)y = x \cos x \)
Solution:
The given equation is in the form \( f(D)y = Q(x) \), where:
\[ f(D) = D^2 + 2D + 1 \quad \text{and} \quad Q(x) = x \cos x \]
The auxiliary equation is:
\[ m^2 + 2m + 1 = 0 \Rightarrow m = -1, -1 \]
Thus, the complementary function (C.F.) is:
\[ \text{C.F.} = (c_1 + c_2 x) e^{-x} \]
The particular integral (P.I.) is:
\[ \text{P.I.} = \frac{1}{D^2 + 2D + 1} x \cos x \]
Using the rule for \( Q(x) = x V(x) \), where \( V(x) = \cos x \):
\[ \text{P.I.} = \left[ x - \frac{1}{f(D)} f'(D) \right] \frac{1}{f(D)} V(x) \]
Here, \( f(D) = D^2 + 2D + 1 \) and \( f'(D) = 2D + 2 \).
Thus:
\[ \text{P.I.} = \left[ x - \frac{1}{D^2 + 2D + 1} (2D + 2) \right] \frac{1}{D^2 + 2D + 1} \cos x \]
Simplify \( \frac{1}{D^2 + 2D + 1} \cos x \):
\[ \frac{1}{D^2 + 2D + 1} \cos x = \frac{1}{-1 + 2D + 1} \cos x = \frac{1}{2D} \cos x \]
Integrate \( \cos x \):
\[ \frac{1}{2D} \cos x = \frac{1}{2} \int \cos x \, dx = \frac{1}{2} \sin x \]
Now, compute \( \frac{1}{D^2 + 2D + 1} (2D + 2) \sin x \):
\[ \frac{1}{D^2 + 2D + 1} (2D + 2) \sin x = \frac{1}{D^2 + 2D + 1} (2 \cos x + 2 \sin x) \]
Simplify further:
\[ \text{P.I.} = \left[ x - \frac{2 \cos x + 2 \sin x}{D^2 + 2D + 1} \right] \frac{1}{2} \sin x \]
Evaluate \( \frac{2 \cos x + 2 \sin x}{D^2 + 2D + 1} \):
\[ \frac{2 \cos x + 2 \sin x}{D^2 + 2D + 1} = \frac{2 \cos x + 2 \sin x}{-1 + 2D + 1} = \frac{2 \cos x + 2 \sin x}{2D} \]
Integrate:
\[ \frac{2 \cos x + 2 \sin x}{2D} = \frac{1}{D} (\cos x + \sin x) = \sin x - \cos x \]
Thus:
\[ \text{P.I.} = \left[ x - (\sin x - \cos x) \right] \frac{1}{2} \sin x = \frac{x}{2} \sin x - \frac{1}{2} (\sin x - \cos x) \sin x \]
Simplify the expression:
\[ \text{P.I.} = \frac{x}{2} \sin x - \frac{1}{2} \sin^2 x + \frac{1}{2} \cos x \sin x \]
The complete solution is:
\[ y = \text{C.F.} + \text{P.I.} = (c_1 + c_2 x) e^{-x} + \frac{x}{2} \sin x + \frac{1}{2} (\cos x - \sin x) \]
2. Solve the Differential Equation \(\left( {{D^2} + 1} \right)y = {x^2}\sin 2x\)
Solution
The auxiliary equation is: \[ m^2 + 1 = 0 \implies m = \pm i \] Therefore, the complementary function (C.F.) is: \[ \text{C.F.} = c_1 \cos x + c_2 \sin x \]
Find the Particular Integral (P.I.)
To find the particular integral, we use the method of undetermined coefficients. The right-hand side of the equation is \(x^2 \sin 2x\), which can be expressed as the imaginary part of \(e^{2ix} x^2\): \[ \text{P.I.} = \text{Imaginary Part of } e^{2ix} \frac{1}{(D + 2i)^2 + 1} x^2 \] Replace \(D\) by \(D + 2i\): \[ \text{P.I.} = \text{Imaginary Part of } e^{2ix} \frac{1}{-3\left[1 - \left(\frac{4iD + D^2}{3}\right)\right]} x^2 \] Expand the denominator using the binomial series: \[ \text{P.I.} = \text{Imaginary Part of } \frac{e^{2ix}}{-3} \left[1 + \left(\frac{4iD + D^2}{3}\right) + \left(\frac{4iD + D^2}{3}\right)^2 + \cdots \right] x^2 \] Simplify the expansion: \[ \text{P.I.} = \text{Imaginary Part of } \frac{e^{2ix}}{-3} \left[1 + \frac{4i}{3}D - \frac{13}{9}D^2 + \cdots \right] x^2 \] Apply the differential operators: \[ \text{P.I.} = \text{Imaginary Part of } \frac{e^{2ix}}{-3} \left[x^2 + \frac{4i}{3}(2x) - \frac{13}{9}(2)\right] \] Simplify further: \[ \text{P.I.} = \text{Imaginary Part of } \left\{ \frac{-1}{3} (\cos 2x + i \sin 2x) \left[\left(x^2 - \frac{26}{9}\right) + i \frac{8x}{3}\right] \right\} \] Extract the imaginary part: \[ \text{P.I.} = \frac{-1}{3} \left[\left(x^2 - \frac{26}{9}\right) \sin 2x + \frac{8}{3} x \cos 2x \right] \]
The Complete Solution
The complete solution is the sum of the complementary function and the particular integral: \[ y = c_1 \cos x + c_2 \sin x - \frac{1}{3} \left[\left(x^2 - \frac{26}{9}\right) \sin 2x + \frac{8}{3} x \cos 2x \right] \]
Model Problems
Solve the following differential equations:
-
Solve: \(\frac{d^2y}{dx^2} + 16y = x \sin 3x\)
Answer: \(y = c_1 \cos 4x + c_2 \sin 4x + \frac{1}{49} (7x \sin 3x - 6 \cos 3x)\) -
Solve: \(\frac{d^2y}{dx^2} - y = x \sin x + x^2 e^x\)
Answer: \(y = c_1 e^x + c_2 e^{-x} - \frac{1}{2} \left[ x \sin x + \cos x \right] + \frac{x e^x}{12} \left( 2x^2 - 3x + 3 \right)\) -
Solve: \(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = x e^x \sin x\)
Answer: \(y = (c_1 + c_2 x) e^{-x} - e^x \left( x \sin x + 2 \cos x \right)\) -
Solve: \((D^2 + 1)y = x \cos x\)
Answer: \(y = c_1 \cos x + c_2 \sin x + \frac{1}{4} \left( x^2 \sin x + x \cos x \right)\) -
Solve: \((D^2 - 1)y = x^2 \cos x\)
Answer: \(y = c_1 e^{-x} + c_2 e^x - \frac{1}{2} \left( x^2 \cos x - 2x \sin x \right)\) -
Solve: \((D^2 - 4D + 4)y = 8x^2 e^{2x} \sin 2x\)
Answer: \(y = (c_1 + c_2 x) e^{2x} + e^{2x} \left[ (3 - 2x^2) \sin 2x - 4x \cos 2x \right]\)