JEE Math Practice Question
The solution of the equation $$\,{{{d^2}y} \over {d{x^2}}} = {e^{ - 2x}}$$
- A.$${{{e^{ - 2x}}} \over 4}$$
- B.$${{{e^{ - 2x}}} \over 4} + cx + d$$
- C.$${1 \over 4}{e^{ - 2x}} + c{x^2} + d$$
- D.$$\,{1 \over 4}{e^{ - 4x}} + cx + d$$
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