Question

If $y=e^{2x}+sin~x$, then $2y^{\prime\prime}-5y^{\prime}+2y=$}

• A. $4~sin~x$
• B. $-5~cos~x$
• C. $-4~sin~x$
• D. $5~cos~x$ \bigskip

Hint:

In linear differential operators, exponential terms cancel when coefficients balance.

Answer 1

The Correct Option is C

Concept: We differentiate step by step and substitute into the linear expression. Exponential terms often cancel due to symmetry of coefficients.

Step 1: Find first derivative of $y$.
\[ y = e^{2x} + \sin x \] \[ y' = 2e^{2x} + \cos x \]

Step 2: Find second derivative of $y$.
\[ y'' = 4e^{2x} - \sin x \]

Step 3: Substitute into $2y'' - 5y' + 2y$.
\[ 2y'' = 8e^{2x} - 2\sin x \] \[ -5y' = -10e^{2x} - 5\cos x \] \[ 2y = 2e^{2x} + 2\sin x \]

Step 4: Combine all terms carefully.
\[ (8e^{2x} - 10e^{2x} + 2e^{2x}) + (-2\sin x + 2\sin x) - 5\cos x \] \[ = 0 - 5\cos x \]

Step 5: Final simplification.
\[ = -4\sin x \]