For a differentiable function f: mathbb{R} ri | vedclass

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(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} - 3y = \alpha$.The integrating factor is $IF = e^{\in

Vedclass · Accepted Answer

$61$

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(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} - 3y = \alpha$.The integrating factor is $IF = e^{\int -3 dx} = e^{-3x}$.Multiplying both sides by $IF$,we get $\frac{d}{dx}(y e^{-3x}) = \alpha e^{-3x}$.Integrating both sides,$y e^{-3x} = \int \alpha e^{-3x} dx = \frac{\alpha e^{-3x}}{-3} + C$.Thus,$y = -\frac{\alpha}{3} + C e^{3x}$.Given $\lim_{x ightarrow -\infty} f(x) = 7$,as $x ightarrow -\infty$,$e^{3x} ightarrow 0$. Therefore,$7 = -\frac{\alpha}{3}$,which implies $\alpha = -21$.Substituting $\alpha = -21$ into the equation,$y = 7 + C e^{3x}$.Using $f(0) = 1$,we get $1 = 7 + C$,so $C = -6$.Thus,$f(x) = 7 - 6 e^{3x}$.Now,calculate $f(-\log_e 3) = 7 - 6 e^{3(-\log_e 3)} = 7 - 6 e^{\log_e(3^{-3})} = 7 - 6(3^{-3}) = 7 - 6(\frac{1}{27}) = 7 - \frac{6}{27} = 7 - \frac{2}{9} = \frac{63-2}{9} = \frac{61}{9}$.Therefore,$9 f(-\log_e 3) = 9 	imes \frac{61}{9} = 61$.

For a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$,suppose $f^{\prime}(x)=3 f(x)+\alpha$,where $\alpha \in \mathbb{R}, f(0)=1$ and $\lim _{x \rightarrow-\infty} f(x)=7$. Then $9 f\left(-\log _{e} 3\right)$ is equal to ............

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