Exams Joint Entrance Examination Mathematics Differential Equations Question 75 Question 75 of 490 Medium NUMERICAL
For a differentiable function f : R → R f:R\to R f : R → R , suppose f ′ ( x ) = 3 f ( x ) + α {f}^{'}(x)=3f(x)+\alpha f ′ ( x ) = 3 f ( x ) + α , where α ∈ R , f ( 0 ) = 1 \alpha \in R,f(0)=1 α ∈ R , f ( 0 ) = 1 and l i m x → − ∞ f ( x ) = 7 \underset{x\to -\infty}{lim}f(x)=7 x → − ∞ l im f ( x ) = 7 . Then 9 f ( − log e 3 ) 9f(-{\log }_{e}3) 9 f ( − log e 3 ) is equal to \_\_\_\_\_\_\_\_\_.
Answer: 61
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f ′ ( x ) = 3 f ( x ) + α ⇒ d y 3 y + α = d x ⇒ 1 3 ln ( 3 y + α ) = x + C y ( 0 ) = 1 ⇒ C = 1 3 ln ( 3 + α ) 1 3 ln ( 3 y + α 3 + α ) = x ⇒ y = 1 3 ( ( 3 + α ) e 3 x − α ) = f ( x ) l i m x → − ∞ f ( x ) = 7 ⇒ α = − 21 ⇒ f ( x ) = 7 − 6 e 3 x 9 f ( − ln 3 ) = 61 \begin{gathered} {f}^{'}(x)=3f(x)+\alpha \\ \Rightarrow \frac{dy}{3y+\alpha }=dx \\ \Rightarrow \frac{1}{3}\ln (3y+\alpha )=x+C \\ y(0)=1\Rightarrow C=\frac{1}{3}\ln (3+\alpha ) \\ \frac{1}{3}\ln (\frac{3y+\alpha }{3+\alpha })=x \\ \Rightarrow y=\frac{1}{3}((3+\alpha ){e}^{3x}-\alpha )=f(x) \\ \underset{x\to -\infty}{lim}f(x)=7\Rightarrow \alpha =-21 \\ \Rightarrow f(x)=7-6{e}^{3x} \\ 9f(-\ln 3)=61 \end{gathered} f ′ ( x ) = 3 f ( x ) + α ⇒ 3 y + α d y = d x ⇒ 3 1 ln ( 3 y + α ) = x + C y ( 0 ) = 1 ⇒ C = 3 1 ln ( 3 + α ) 3 1 ln ( 3 + α 3 y + α ) = x ⇒ y = 3 1 (( 3 + α ) e 3 x − α ) = f ( x ) x → − ∞ l im f ( x ) = 7 ⇒ α = − 21 ⇒ f ( x ) = 7 − 6 e 3 x 9 f ( − ln 3 ) = 61 Answer the question to unlock