Question

Let $\frac{dy}{dx}+y=f\left(x\right)$ where $y$ is a continuous function of $x$ with $y\left(0\right)=1$ and $f\left(x\right)=\{\begin{array}{cc}{e}^{-x},& \text{if}0\le x\le 2\\ e^{-2},& \text{if}x>2\end{array}$
Which of the following hold(s) good ?

• A. $y\left(1\right)=2e^{-1}$
• B. $y^{\prime }\left(1\right)=-e^{-1}$
• C. $y\left(3\right)=-2e^{-3}$
• D. $y^{\prime }\left(3\right)=-2e^{-3}$

Answer 1

Answer: (A), (B), (D)

$y^{\prime }\left(3\right)=-2e^{-3}$

$\frac{dy}{dx}+y=f\left(x\right)$
I. F. $=e^x$
Solution is $y{e}^x=\int e^{x}f\left(x\right)dx+C$

Now, if $0\le x\le 2,$ then
$y{e}^x=\int e^x{e}^{-x}dx+C$
$\Rightarrow y{e}^x=x+C$
$x=0,y\left(0\right)=1\Rightarrow C=1$
$\therefore y{e}^x=x+1$
$\Rightarrow y\left(x\right)=\frac{x+1}{e^x}$
$y\left(1\right)=\frac{2}{e}$

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

If $x>2,$ then
$y{e}^x=\int e^{x-2}dx$
$\Rightarrow y{e}^x=e^{x-2}+C$
$\Rightarrow y=e^{-2}+C{e}^{-x}$

As $y$ is continuous,
$\therefore \underset{x\to 2}{lim}\frac{x+1}{e^x}=\underset{x\to 2}{\text{lim}}(e^{-2}+C{e}^{-x})$
$\Rightarrow 3e^{-2}=e^{-2}+C{e}^{-2}$
$\Rightarrow C=2$

$\therefore$ For $x>2,$
$y\left(x\right)=e^{-2}+2e^{-x}$
Hence, $y\left(3\right)=2e^{-3}+e^{-2}=e^{-2}(2e^{-1}+1)$
$y^{\prime }\left(x\right)=-2e^{-x}$
$y^{\prime }\left(3\right)=-2e^{-3}$