Question

The particular integral for the differential equation \( (D^2 - 2D + 1)y = x^2 e^{3x} \) is

• A. \( \frac{1}{8} e^{3x} (2x^2 + 4x - 3) \)
• B. \( \frac{1}{8} e^{3x} (2x^2 + 4x + 3) \)
• C. \( \frac{1}{8} e^{3x} (2x^2 - 4x + 3) \)
• D. \( \frac{1}{8} e^{3x} (2x^2 - 4x - 3) \)

Hint:

Particular Integral (Operator Method). For \(F(D)y = e^{axV(x)\), use shift theorem: \(y_p = e^{ax \frac{1{F(D+a) V(x)\). If V(x) is a polynomial, expand \(1/F(D+a)\) using binomial series and operate on V(x).

Answer 1

The Correct Option is C

The differential equation is \( (D^2 - 2D + 1)y = x^2 e^{3x} \), which is \( (D-1)^2 y = x^2 e^{3x} \).
We need to find the particular integral (\(y_p\)).
Using the operator method: $$ y_p = \frac{1}{(D-1)^2} (x^2 e^{3x}) $$ We use the shift theorem: \( \frac{1}{F(D)} (e^{ax} V) = e^{ax} \frac{1}{F(D+a)} V \).
Here, \(a=3\) and \(V=x^2\).
\(F(D) = (D-1)^2\).
$$ y_p = e^{3x} \frac{1}{((D+3)-1)^2} x^2 = e^{3x} \frac{1}{(D+2)^2} x^2 $$ Now we need to evaluate \( \frac{1}{(D+2)^2} x^2 \).
We can use binomial expansion: $$ \frac{1}{(D+2)^2} = \frac{1}{[2(1+D/2)]^2} = \frac{1}{4} (1+D/2)^{-2} $$ Using the expansion \( (1+u)^{-2} = 1 - 2u + 3u^2 - \dots \): $$ \frac{1}{4} (1 - 2(D/2) + 3(D/2)^2 - \dots) = \frac{1}{4} (1 - D + \frac{3}{4}D^2 - \dots) $$ Apply this operator to \(x^2\): $$ y_p = e^{3x} \frac{1}{4} (1 - D + \frac{3}{4}D^2 - \dots) x^2 $$ $$ D(x^2) = 2x $$ $$ D^2(x^2) = 2 $$ $$ D^3(x^2) = 0 $$ So we only need terms up to \(D^2\).
$$ y_p = \frac{e^{3x}}{4} [1(x^2) - D(x^2) + \frac{3}{4}D^2(x^2)] $$ $$ y_p = \frac{e^{3x}}{4} [x^2 - 2x + \frac{3}{4}(2)] $$ $$ y_p = \frac{e^{3x}}{4} [x^2 - 2x + \frac{3}{2}] $$ To match the options which have a factor of 1/8, multiply inside by 2 and outside by 1/2: $$ y_p = \frac{e^{3x}}{8} [2x^2 - 4x + 3] $$ This matches option (3).