Differential Equations Question 3
Question 3 - 2024 (01 Feb Shift 2)
Let $\alpha$ be a non-zero real number. Suppose $f: \mathrm{R} \rightarrow \mathrm{R}$ is a differentiable function such that $f(0)=2$ and $\lim _{\mathrm{x} \rightarrow-\infty} \mathrm{f}(\mathrm{x})=1$. If $f^{\prime}(\mathrm{x})=\alpha f(x)+3$, for all $\mathrm{x} \in \mathrm{R}$, then $f\left(-\log _{e} 2\right)$ is equal to
(1) 3
(2) 5
(3) 9
(4) 7
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Answer (3)
Solution
$f(0)=2, \lim _{x \rightarrow-\infty} f(x)=1$
$f^{\prime}(x)-x \cdot f(x)=3$
I.F $=e^{-\alpha x}$
$y\left(e^{-\alpha x}\right)=\int 3 \cdot e^{-\alpha x} d x$
$f(x) \cdot\left(e^{-\alpha x}\right)=\frac{3 e^{-\alpha x}}{-\alpha}+c$
$x=0 \Rightarrow 2=\frac{-3}{\alpha}+c \Rightarrow \frac{3}{\alpha}=c-2 \ldots$
$f(x)=\frac{-3}{\alpha}+c \cdot e^{\alpha x}$
$x \rightarrow-\infty \Rightarrow 1=\frac{-3}{\alpha}+c(0)$
$\alpha=-3 \therefore c=1$
$f(-\ln 2)=\frac{-3}{\alpha}+c \cdot e^{\alpha x}$
$=1+e^{3 \ln 2}=9$
(But $\alpha$ should be greater than 0 for finite value of $\mathrm{c}$ )
