Let alpha be a non-zero real number. Suppose | vedclass

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(A) Given the linear differential equation $f'(x) - \alpha f(x) = 3$.
The integrating factor is $I.F. = e^{\int -\alpha dx} = e^{-\alpha x}$.
Mu

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$2$

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(A) Given the linear differential equation $f'(x) - \alpha f(x) = 3$.
The integrating factor is $I.F. = e^{\int -\alpha dx} = e^{-\alpha x}$.
Multiplying both sides by the $I.F.$,we get $\frac{d}{dx} [f(x) e^{-\alpha x}] = 3 e^{-\alpha x}$.
Integrating both sides,$f(x) e^{-\alpha x} = \int 3 e^{-\alpha x} dx = -\frac{3}{\alpha} e^{-\alpha x} + C$.
Thus,$f(x) = -\frac{3}{\alpha} + C e^{\alpha x}$.
Given $f(0) = 2$,we have $2 = -\frac{3}{\alpha} + C$,so $C = 2 + \frac{3}{\alpha}$.
Given $\lim_{x \rightarrow -\infty} f(x) = 1$.
If $\alpha > 0$,then $e^{\alpha x} \rightarrow 0$ as $x \rightarrow -\infty$,so $f(x) \rightarrow -\frac{3}{\alpha} = 1$,which implies $\alpha = -3$. This contradicts $\alpha > 0$.
If $\alpha < 0$,then $e^{\alpha x} \rightarrow \infty$ as $x \rightarrow -\infty$. For the limit to be $1$,the coefficient of $e^{\alpha x}$ must be $0$.
So $C = 0$,which means $2 + \frac{3}{\alpha} = 0$,so $\alpha = -\frac{3}{2}$.
Then $f(x) = -\frac{3}{-3/2} = 2$. Since $f(x) = 2$ is a constant function,$f'(x) = 0$. The equation $f'(x) = \alpha f(x) + 3$ becomes $0 = (-3/2)(2) + 3 = 0$,which is consistent.
Thus $f(x) = 2$ for all $x$.
Therefore,$f(-\log_e 2) = 2$.

Let $\alpha$ be a non-zero real number. Suppose $f: R \rightarrow R$ is a differentiable function such that $f(0)=2$ and $\lim _{x \rightarrow-\infty} f(x)=1$. If $f^{\prime}(x)=\alpha f(x)+3$ for all $x \in R$,then $f(-\log _e 2)$ is equal to . . . . . . . . .

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