MCQ
If the function $y = e^{4x} + 2e^{-x}$ is a solution of the differential equation $\frac{{\frac{{{d^3}y}}{{d{x^3}}} - 13\frac{{dy}}{{dx}}}}{y} = K$ then the value of $K$ is
- A$4$
- B$6$
- C$9$
- ✓$12$
✓
Answer
Correct option: D.
$12$
d
$y = e^{4x} + 2e^{-x} ; \,\, y_1 = 4e^{4x} - 2e^{-x} ;$
$y_2 = 16e^{4x} + 2e^{-x} ; \,\,y_3 = 64e^{4x} - 2e^{-x}$
Now, $y_3 - 13y_1 = (64e^{4x} - 2e^{-x}) - 13(4e^{4x} - 2e^{-x}) = 12e^{4x} + 24e^{-x}$
$= 12(e^{4x} + 2e^{-x}) = 12y$
$\frac{{{y_3} - 13{y_1}}}{y} = 12$
$y = e^{4x} + 2e^{-x} ; \,\, y_1 = 4e^{4x} - 2e^{-x} ;$
$y_2 = 16e^{4x} + 2e^{-x} ; \,\,y_3 = 64e^{4x} - 2e^{-x}$
Now, $y_3 - 13y_1 = (64e^{4x} - 2e^{-x}) - 13(4e^{4x} - 2e^{-x}) = 12e^{4x} + 24e^{-x}$
$= 12(e^{4x} + 2e^{-x}) = 12y$
$\frac{{{y_3} - 13{y_1}}}{y} = 12$
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