Question

Which of the following is NOT a solution of the differential equation \(\dfrac{d^2y}{dx^2}+y=1\)?

• A. \(y=1\)
• B. \(y=1+\cos x\)
• C. \(y=1+\sin x\)
• D. \(y=2+\sin x+\cos x\)

Hint:

For \(y''+y=1\), the general solution is \(y=1+c_1\cos x+c_2\sin x\).

Answer 1

The Correct Option is D

Concept:
To check whether a function is a solution of a differential equation, substitute the function and its derivatives into the equation. The given differential equation is \[ \frac{d^2y}{dx^2}+y=1 \]

Step 1: Find the general nature of the solution. The complementary equation is \[ \frac{d^2y}{dx^2}+y=0 \] Its auxiliary equation is \[ m^2+1=0 \] \[ m=\pm i \] So the complementary function is \[ c_1\cos x+c_2\sin x \]

Step 2: Find a particular solution. Since the right side is \(1\), take a constant particular solution: \[ y_p=1 \] Then, \[ y_p''=0 \] and \[ y_p''+y_p=0+1=1 \] So \(y_p=1\) is valid.

Step 3: Write the general solution. \[ y=1+c_1\cos x+c_2\sin x \] Therefore, any solution must have constant term \(1\).

Step 4: Check the options. \[ y=1 \] is a solution. \[ y=1+\cos x \] is a solution. \[ y=1+\sin x \] is also a solution. But \[ y=2+\sin x+\cos x \] has constant term \(2\), not \(1\).

Step 5: Direct verification of option (D). If \[ y=2+\sin x+\cos x \] then \[ y''=-\sin x-\cos x \] So, \[ y''+y=(-\sin x-\cos x)+(2+\sin x+\cos x) \] \[ y''+y=2 \] But the equation requires \[ y''+y=1 \] Hence option (D) is not a solution.

Step 6: Final answer. \[ \boxed{y=2+\sin x+\cos x} \]