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

Concept: The given linear differential equation can be written using differential operator notation \( D = \frac{d}{dx} \): \[ (D^2 - 4D)y = x^2 e^{2x} \] The particular integral \( y_p \) is found using the inverse operator: \[ y_p = \frac{1}{D^2 - 4D} \left( x^2 e^{2x} \right) \] To evaluate this when an exponential function \( e^{ax} \) is multiplied by another function, we use the exponential shift rule: \[ \frac{1}{f(D)} \left( e^{ax} V(x) \right) = e^{ax} \frac{1}{f(D+a)} V(x) \]

Step 1: Setting up the particular integral operator expression.
Identify our function of the differential operator from the differential equation: \[ f(D) = D^2 - 4D \] Here, our exponential multiplier term has a power coefficient of \( a = 2 \), and the remaining function is \( V(x) = x^2 \). Applying the shift rule changes the operator \( D \rightarrow D + 2 \): \[ y_p = e^{2x} \frac{1}{(D+2)^2 - 4(D+2)} (x^2) \]

Step 2: Simplifying the shifted denominator operator.
Let us expand the algebraic terms in the denominator: \[ (D+2)^2 = D^2 + 4D + 4 \] \[ -4(D+2) = -4D - 8 \] Adding these expressions together: \[ f(D+2) = (D^2 + 4D + 4) + (-4D - 8) = D^2 - 4 \] Substitute this back into the expression for \( y_p \): \[ y_p = e^{2x} \frac{1}{D^2 - 4} (x^2) \quad \cdots (1) \]

Step 3: Expanding the operator via binomial expansion for polynomial terms.
To apply the operator to a polynomial like \( x^2 \), we rewrite the denominator in the form \( -(4 - D^2) \) and factor out the constant to use a binomial series: \[ \frac{1}{D^2 - 4} = \frac{1}{-4 \left(1 - \frac{D^2}{4}\right)} = -\frac{1}{4} \left(1 - \frac{D^2}{4}\right)^{-1} \] Using the binomial expansion formula \( (1-t)^{-1} = 1 + t + t^2 + \cdots \): \[ \left(1 - \frac{D^2}{4}\right)^{-1} = 1 + \frac{D^2}{4} + \frac{D^4}{16} + \cdots \] Since our target polynomial expression is \( x^2 \), any derivative higher than the second derivative will equal zero. Therefore, we can drop terms containing \( D^4 \) and higher powers: \[ y_p = e^{2x} \left[ -\frac{1}{4} \left( 1 + \frac{D^2}{4} \right) \right] (x^2) \]

Step 4: Distributing the operator onto the polynomial \( x^2 \).
Now, let's apply the operations inside the brackets to the polynomial: \[ y_p = -\frac{1}{4} e^{2x} \left[ x^2 + \frac{1}{4} D^2(x^2) \right] \] Compute the required derivatives of \( x^2 \): \[ D(x^2) = \frac{d}{dx}(x^2) = 2x \] \[ D^2(x^2) = \frac{d}{dx}(2x) = 2 \] Substitute \( D^2(x^2) = 2 \) back into the expression: \[ y_p = -\frac{1}{4} e^{2x} \left( x^2 + \frac{1}{4} \cdot 2 \right) = -\frac{1}{4} e^{2x} \left( x^2 + \frac{1}{2} \right) \]

Step 5: Equating to find the target function \( y(x) \).
The problem states that the particular integral is written in the form \( y_p = e^{2x} y(x) \). Comparing this directly with our calculated result: \[ e^{2x} y(x) = e^{2x} \left[ -\frac{1}{4} \left( x^2 + \frac{1}{2} \right) \right] \] Dividing out the matching exponential factor from both sides isolates our target function: \[ y(x) = -\frac{1}{4} \left( x^2 + \frac{1}{2} \right) \] This matches option (A).