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Find-a-particular-solution-of-the-equation-f-x-f-x-such-that-f-x-2-for-x-2-




Question Number 130893 by EDWIN88 last updated on 30/Jan/21
Find a particular solution  of the equation f ′(x) = f(x)   such that f(x)=2 for x=2
$${Find}\:{a}\:{particular}\:{solution} \\ $$$${of}\:{the}\:{equation}\:{f}\:'\left({x}\right)\:=\:{f}\left({x}\right)\: \\ $$$${such}\:{that}\:{f}\left({x}\right)=\mathrm{2}\:{for}\:{x}=\mathrm{2}\: \\ $$
Answered by mr W last updated on 30/Jan/21
y′=y  (dy/y)=dx  ln y=x+C  ⇒y=ce^x   ⇒f(2)=ce^2 =2 ⇒c=2e^(−2)   ⇒f(x)=y=2e^(x−2)
$${y}'={y} \\ $$$$\frac{{dy}}{{y}}={dx} \\ $$$$\mathrm{ln}\:{y}={x}+{C} \\ $$$$\Rightarrow{y}={ce}^{{x}} \\ $$$$\Rightarrow{f}\left(\mathrm{2}\right)={ce}^{\mathrm{2}} =\mathrm{2}\:\Rightarrow{c}=\mathrm{2}{e}^{−\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)={y}=\mathrm{2}{e}^{{x}−\mathrm{2}} \\ $$
Answered by benjo_mathlover last updated on 30/Jan/21
 y′−y = 0 ⇒λ−1=0 ; λ=1    y(x) = C e^x  ⇒y(2)= Ce^2  = 2 ; C = 2e^(−2)    ∴ y(x) = 2e^(−2) .e^x  = 2e^(x−2)
$$\:\mathrm{y}'−\mathrm{y}\:=\:\mathrm{0}\:\Rightarrow\lambda−\mathrm{1}=\mathrm{0}\:;\:\lambda=\mathrm{1}\: \\ $$$$\:\mathrm{y}\left(\mathrm{x}\right)\:=\:\mathrm{C}\:\mathrm{e}^{\mathrm{x}} \:\Rightarrow\mathrm{y}\left(\mathrm{2}\right)=\:\mathrm{Ce}^{\mathrm{2}} \:=\:\mathrm{2}\:;\:\mathrm{C}\:=\:\mathrm{2e}^{−\mathrm{2}} \: \\ $$$$\therefore\:\mathrm{y}\left(\mathrm{x}\right)\:=\:\mathrm{2e}^{−\mathrm{2}} .\mathrm{e}^{\mathrm{x}} \:=\:\mathrm{2e}^{\mathrm{x}−\mathrm{2}} \\ $$

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