Particular Integral

Case IV: Particular Integral for Exponential-Trigonometric/Polynomial Functions When \( Q(x) = e^{ax} V(x) \)

Consider the differential equation:

\[ f(D)y = Q(x) \]

where \( Q(x) = e^{ax} V(x) \), and \( V(x) \) is either \( \sin bx \), \( \cos bx \), or \( x^k \). Here, \( a \) can be real or complex.

The particular integral (P.I.) is given by:

\[ \text{P.I.} = \frac{1}{f(D)} Q(x) = \frac{1}{f(D)} e^{ax} V(x) \]

To evaluate this, take \( e^{ax} \) out of the operator and replace \( D \) with \( (D + a) \):

\[ \text{P.I.} = e^{ax} \frac{1}{f(D + a)} V(x) \]

Now, operate \( \frac{1}{f(D + a)} \) on \( V(x) \) using the methods discussed earlier.

Steps to Find P.I.:

Take \( e^{ax} \) out of the operator \( \frac{1}{f(D)} \).
Replace every \( D \) in \( f(D) \) with \( (D + a) \) to get \( f(D + a) \).
Operate \( \frac{1}{f(D + a)} \) on \( V(x) \) using the appropriate method (e.g., for trigonometric or polynomial functions).

Example:

If \( Q(x) = e^{2x} \sin 3x \), then:

\[ \text{P.I.} = \frac{1}{f(D)} e^{2x} \sin 3x = e^{2x} \frac{1}{f(D + 2)} \sin 3x \]

Now, evaluate \( \frac{1}{f(D + 2)} \sin 3x \) using the method for trigonometric functions.

1. Solve the differential equation: \( y'' - 2y' + 2y = e^x \cos x \)

Solution:

The given equation can be written as:

\[ (D^2 - 2D + 2)y = e^x \cos x \]

Here, \( f(D) = D^2 - 2D + 2 \) and \( Q(x) = e^x \cos x \).

The auxiliary equation is:

\[ m^2 - 2m + 2 = 0 \Rightarrow m = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i \]

Thus, the complementary function (C.F.) is:

\[ \text{C.F.} = e^x (c_1 \cos x + c_2 \sin x) \]

The particular integral (P.I.) is:

\[ \text{P.I.} = \frac{1}{D^2 - 2D + 2} e^x \cos x \]

Using the rule for \( Q(x) = e^{ax} V(x) \), take \( e^x \) out of the operator and replace \( D \) with \( (D + 1) \):

\[ \text{P.I.} = e^x \frac{1}{(D + 1)^2 - 2(D + 1) + 2} \cos x \]

Simplify the operator:

\[ (D + 1)^2 - 2(D + 1) + 2 = D^2 + 2D + 1 - 2D - 2 + 2 = D^2 + 1 \]

Thus:

\[ \text{P.I.} = e^x \frac{1}{D^2 + 1} \cos x \]

Since \( D^2 + 1 \) applied to \( \cos x \) gives zero, we use the rule for repeated roots:

\[ \text{P.I.} = e^x \cdot x \cdot \frac{1}{2D} \cos x \]

Integrate \( \cos x \):

\[ \text{P.I.} = \frac{x}{2} e^x \int \cos x \, dx = \frac{x}{2} e^x \sin x \]

The complete solution is:

\[ y = \text{C.F.} + \text{P.I.} = e^x (c_1 \cos x + c_2 \sin x) + \frac{x}{2} e^x \sin x \]

2. Solve the differential equation: \( (D^2 + 1)y = \sin x \sin 2x + e^x x^2 \)

Solution:

The given equation is in the form \( f(D)y = Q(x) \), where:

\[ f(D) = D^2 + 1 \quad \text{and} \quad Q(x) = \sin x \sin 2x + e^x x^2 \]

The auxiliary equation is:

\[ m^2 + 1 = 0 \Rightarrow m = \pm i \]

Thus, the complementary function (C.F.) is:

\[ \text{C.F.} = c_1 \cos x + c_2 \sin x \]

The particular integral (P.I.) is split into two parts:

\[ \text{P.I.} = \text{P.I.}_1 + \text{P.I.}_2 \]

1. Calculation of \( \text{P.I.}_1 \):

For \( \text{P.I.}_1 \):

\[ \text{P.I.}_1 = \frac{1}{D^2 + 1} \sin x \sin 2x \]

Use the identity \( \sin x \sin 2x = \frac{1}{2} (\cos x - \cos 3x) \):

\[ \text{P.I.}_1 = \frac{1}{2} \cdot \frac{1}{D^2 + 1} (\cos x - \cos 3x) \]

Evaluate each term separately:

\[ \frac{1}{D^2 + 1} \cos x = \frac{x}{2} \sin x \]

\[ \frac{1}{D^2 + 1} \cos 3x = \frac{1}{-9 + 1} \cos 3x = -\frac{1}{8} \cos 3x \]

Thus:

\[ \text{P.I.}_1 = \frac{1}{2} \left( \frac{x}{2} \sin x - \frac{1}{8} \cos 3x \right) = \frac{x}{4} \sin x - \frac{1}{16} \cos 3x \]

2. Calculation of \( \text{P.I.}_2 \):

For \( \text{P.I.}_2 \):

\[ \text{P.I.}_2 = \frac{1}{D^2 + 1} e^x x^2 \]

Using the rule for \( Q(x) = e^{ax} V(x) \), take \( e^x \) out of the operator and replace \( D \) with \( (D + 1) \):

\[ \text{P.I.}_2 = e^x \frac{1}{(D + 1)^2 + 1} x^2 = e^x \frac{1}{D^2 + 2D + 2} x^2 \]

Rewrite \( \frac{1}{D^2 + 2D + 2} \) as \( \frac{1}{2(1 + \frac{D^2 + 2D}{2})} \):

\[ \text{P.I.}_2 = \frac{e^x}{2} \left( 1 + \frac{D^2 + 2D}{2} \right)^{-1} x^2 \]

Expand \( \left( 1 + \frac{D^2 + 2D}{2} \right)^{-1} \) using the binomial expansion:

\[ \left( 1 + \frac{D^2 + 2D}{2} \right)^{-1} = 1 - \frac{D^2 + 2D}{2} + \left( \frac{D^2 + 2D}{2} \right)^2 - \dots\]

Apply the expansion to \( x^2 \):

\[ \text{P.I.}_2 = \frac{e^x}{2} \left( 1 - \frac{D^2 + 2D}{2} \right) x^2\]

Compute the derivatives:

\[ D(x^2) = 2x\]

\[ D^2(x^2) = 2\]

Substitute back:

\[ \text{P.I.}_2 = \frac{e^x}{2} \left( x^2 - \frac{2 + 4x}{2} \right) = \frac{e^x}{2} (x^2 - 2x - 1)\]

Combine both parts of the particular integral:

\[ \text{P.I.} = \text{P.I.}_1 + \text{P.I.}_2 = \frac{x}{4} \sin x - \frac{1}{16} \cos 3x + \frac{e^x}{2} (x^2 - 2x - 1)\]