Let dy - dx - y - f x where y is a continuous function of x with y 0-1 and - f x -e -2- - x lgt 2 lend cases - Which of the following holds good- A- y1-2 e-1B- y -1- e -1C- y 3-2 e -3D- y -3-2 e -3

The correct options are A y(1)=2e^ 1 B y'(1)= e^ 1 D y'(3)= 2e^ 3dydx+y=f(x) I.F. =e^x nbsp;ye^x=∫ e^xf(x)dx+C Now, if nbsp;0≤ x ≤ 2 then nbsp; ye^x=∫ e^xe ...

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Byju's Answer

Standard XII

Mathematics

Linear Differential Equations of First Order

Let dy / dx +...

Question

Let $\frac{d y}{d x} + y = f (x)$ where $y$ is a continuous function of $x$ with $y (0) = 1$ and $f (x) = {\begin{matrix} e^{- x}, & 0 \leq x \leq 2 e^{- 2}, & x > 2 \end{matrix}$ .
Which of the following hold(s) good?

y (1) = 2 e^{- 1}

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y^{'} (1) = - e^{- 1}

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y (3) = - 2 e^{- 3}

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y^{'} (3) = - 2 e^{- 3}

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Solution

The correct options are
A $y (1) = 2 e^{- 1}$
B $y^{'} (1) = - e^{- 1}$
D $y^{'} (3) = - 2 e^{- 3}$
$\frac{d y}{d x} + y = f (x)$
$I . F . = e^{x} y e^{x} = \int e^{x} f (x) d x + C$

Now, if $0 \leq x \leq 2$
then $y e^{x} = \int e^{x} e^{- x} d x + C$
$\Rightarrow y e^{x} = x + C$
$y (0) = 1 \Rightarrow C = 1$
$∴ y e^{x} = x + 1 y = \frac{x + 1}{e^{x}}, y (1) = \frac{2}{e}$

$y^{'} = \frac{e^{x} - (x + 1) e^{x}}{e^{2 x}} y^{'} (1) = \frac{e - 2 e}{e^{2}} = \frac{- e}{e^{2}} = - \frac{1}{e}$

If $x > 2,$
$y e^{x} = \int e^{x - 2} + C$
$\Rightarrow y = e^{- 2} + C e^{- x}$
As $y$ is continuous,
$lim x \to 2 \frac{x + 1}{e^{x}} = lim x \to 2 (e^{- 2} + C e^{- x})$
$\Rightarrow C = 2$
Hence, $y = e^{- 2} + 2 e^{- x}$
$y (3) = e^{- 2} + 2 e^{- 3}$
$y^{'} = - 2 e^{- x}$
$y^{'} (3) = - 2 e^{- 3}$

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Let dydx+y=f(x) where y is a continuous function of x with y(0)=1 and f(x)={e−x,0≤x≤2e−2,x>2. Which of the following hold(s) good?

Let $\frac{d y}{d x} + y = f (x)$ where $y$ is a continuous function of $x$ with $y (0) = 1$ and $f (x) = {\begin{matrix} e^{- x}, & 0 \leq x \leq 2 e^{- 2}, & x > 2 \end{matrix}$ .
Which of the following hold(s) good?