MCQ
The solution of the equation $\frac{{{d^2}y}}{{d{x^2}}} = {e^{ - 2x}}$ is
- A$\frac{1}{4}{e^{ - 2x}}$
- ✓$\frac{1}{4}{e^{ - 2x}} + cx + d$
- C$\frac{1}{4}{e^{ - 2x}} + c{x^2} + d$
- D$\frac{1}{4}{e^{ - 2x}} + c + d$
✓
Answer
Correct option: B.
$\frac{1}{4}{e^{ - 2x}} + cx + d$
b
(b) $\frac{{{d^2}y}}{{d{x^2}}} = {e^{ - 2x}}$
(b) $\frac{{{d^2}y}}{{d{x^2}}} = {e^{ - 2x}}$
Integrating both sides, we get $\frac{{dy}}{{dx}} = \frac{{{e^{ - 2x}}}}{{ - 2}} + c$
Again integrate, we get $y = \frac{{{e^{ - 2x}}}}{4} + cx + d$.
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