Question

If the particular integral of \( y'' - 4y' = x^2 e^{2x} \) is in the form \( y_p = e^{2x} y(x) \), then \( y(x) \) is

• A. \( -\frac{1}{4}\left(x^2 + \frac{1}{2}\right) \)
• B. \( \frac{1}{4}\left(x^2 + \frac{1}{2}\right) \)
• C. \( x^2 + \frac{1}{2} \)
• D. \( x + \frac{1}{2} \)

Hint:

Always ensure the constant term is factored out as a positive 1 before running your binomial expansion series. Forgetting to factor out the negative sign from \( D^2-4 \) is a common trap that leads to incorrect signs!

Answer 1

The Correct Option is A

Step 1: Substitute \( y = e^{2x}u(x) \) directly into the equation.
Instead of using the operator shift rule, plug the assumed form straight into \( y'' - 4y' = x^2e^{2x} \). With \( y = e^{2x}u \): \[ y' = e^{2x}(2u + u'), \qquad y'' = e^{2x}(4u + 4u' + u'') \] So \[ y'' - 4y' = e^{2x}\left[(4u+4u'+u'') - 4(2u+u')\right] = e^{2x}(u'' - 4u) \] Matching this to the right side \( x^2e^{2x} \) gives a much simpler equation for \( u \): \[ u'' - 4u = x^2 \]
Step 2: Try a quadratic for \( u \).
Since the right side is a plain quadratic, try \( u = Ax^2 + Bx + C \). Then \( u' = 2Ax + B \) and \( u'' = 2A \). Substituting: \[ 2A - 4(Ax^2 + Bx + C) = x^2 \] \[ -4Ax^2 - 4Bx + (2A - 4C) = x^2 + 0\cdot x + 0 \]
Step 3: Match coefficients.
Comparing powers of \( x \) on both sides: \[ -4A = 1 \Rightarrow A = -\frac{1}{4}, \qquad -4B = 0 \Rightarrow B = 0, \qquad 2A - 4C = 0 \Rightarrow C = \frac{A}{2} = -\frac{1}{8} \] So \[ u(x) = -\frac{1}{4}x^2 - \frac{1}{8} = -\frac{1}{4}\left(x^2 + \frac{1}{2}\right) \] Since \( y_p = e^{2x}u(x) \), we get \[ \boxed{y(x) = -\frac{1}{4}\left(x^2 + \frac{1}{2}\right)} \]
which matches option (A).