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 $f\left(x\right)=\{\begin{array}{cc}{e}^{-x},& 0\le x\le 2\\ e^{-2},& x>2\end{array}.$
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)

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

$\frac{dy}{dx}+y=f\left(x\right)$ is first order linear differential equation.
I.F. $=e^x$
General solution is $y{e}^x=\int e^{x}f\left(x\right)dx+C$

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


If $x>2,$ then $f\left(x\right)=e^{-2}$
$\therefore y{e}^x=\int e^x\cdot e^{-2}dx+C$
$\Rightarrow y=e^{-2}+C{e}^{-x}$
Since $y$ is a continuous function,
$\underset{x\to 2}{lim}{e}^{-x}(x+1)=\underset{x\to 2}{lim}(e^{-2}+C{e}^{-x})$
$\Rightarrow 3e^{-2}=e^{-2}(C+1)$
$\Rightarrow C=2$
$\therefore y=e^{-2}+2e^{-x}$
$y^{\prime }=-2e^{-x}$
Hence, $y\left(3\right)=e^{-2}+2e^{-3}$ and $y^{\prime }\left(3\right)=-2e^{-3}$