JEE Mains · Maths · STD 12 - 9. differential equations
For a differentiable function \(\mathrm{f}: I R \rightarrow I R\), suppose \(f^{\prime}(\mathrm{x})=3 f(\mathrm{x})+\alpha\), where \(\alpha \in \operatorname{IR}, f(0)=1\) and \(\lim _{x \rightarrow-\infty} f(x)=7\). Then \(9 \mathrm{f}\left(-\log _{\mathrm{e}} 3\right)\) is equal to ............
- A \(12\)
- B \(87\)
- C \(61\)
- D \(25\)
Answer & Solution
Correct Answer
(C) \(61\)
Step-by-step Solution
Detailed explanation
\( \frac{d y}{d x}-3 y=\alpha \) \( \text { If }=e^{\int-3 d x}=e^{-3 x} \) \( \therefore y-e^{-3 x}=\int e^{-3 x} \cdot \alpha d x \) \( y e^{-3 x}=\frac{\alpha e^{-3 x}}{-3}+c \) \( \left(* e^{3 x}\right) \) \( y=\frac{\alpha}{-3}+C \cdot e^{3 x}\) on substituting…
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