Question

For an ordinary differential equation \( y'' + y' = x^2 + 2x + 4 \), the particular integral is given by:

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

Hint:

For solving non-homogeneous differential equations, assume a particular solution based on the form of the non-homogeneous term.

Answer 1

The Correct Option is D

Step 1: Solving the differential equation.
We start with the differential equation: \[ y'' + y' = x^2 + 2x + 4 \] First, we find the particular solution. We assume a trial solution of the form: \[ y_p = Ax^3 + Bx^2 + Cx + D \]

Step 2: Find the derivatives.
Now, differentiate \(y_p\): \[ y_p' = 3Ax^2 + 2Bx + C \] \[ y_p'' = 6Ax + 2B \]

Step 3: Substitute into the equation.
Substitute \(y_p'\) and \(y_p''\) into the original equation: \[ (6Ax + 2B) + (3Ax^2 + 2Bx + C) = x^2 + 2x + 4 \]

Step 4: Simplify and equate coefficients.
By equating the coefficients of like powers of \(x\), we get: - \(3A = 1\) (coefficient of \(x^2\)) \(\Rightarrow A = \frac{1}{3}\) - \(6A + 2B = 2\) (coefficient of \(x\)) \(\Rightarrow 2B = -2 $\Rightarrow$ B = -1\) - \(2B + C = 4\) (constant term) \(\Rightarrow C = 6\) Thus, the particular integral is: \[ y_p = \frac{x^3}{3} - x^2 + 6x \]

Final Answer: \[ \boxed{\frac{x^3}{3} + 2x} \]