A) -13, 14
B) -13, -12
C) -13, 12
D) 12, -13
Correct Answer: B
Solution :
[b] Given \[y={{e}^{4x}}+2{{e}^{-x}}\] Differentiating we get \[\frac{dy}{dx}=4{{e}^{4x}}-2{{e}^{-x}}\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=16{{e}^{14x}}+2{{e}^{-x}}\] \[\Rightarrow \frac{{{d}^{3}}y}{d{{x}^{3}}}=64{{e}^{4x}}-2{{e}^{-x}}\] Putting these values in \[\frac{{{d}^{3}}y}{d{{x}^{3}}}+A\frac{dy}{dx}+By=0\] We have, \[(64+4A+B){{e}^{4x}}+(-2-2A+2B){{e}^{-x}}=0\] \[\Rightarrow \,\,\,64+4A+B=0,\,\,\,-2-2A+2B=0\] Solving these eqs. we get \[A=-13,B=-12\]You need to login to perform this action.
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