# Problem of the Week #38 - December 17th, 2012

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#### Chris L T521

##### Well-known member
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Thanks to those who participated in last week's POTW!! Here's this week's problem!

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Problem: Find a continuous solution to the following differential equation:

$\frac{dy}{dx}+2y=f(x);\qquad f(x)=\begin{cases}1, & 0\leq x\leq 3\\ 0, & x>3\end{cases}$

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#### Chris L T521

##### Well-known member
Staff member
This week's question was correctly answered by Sudharaka and Deveno (MarkFL gets honorable mention - most of his solution was correct, but missed a thing or two). You can find Sudharaka's solution below.

When $$0\leq x\leq 3$$ we have,

$\frac{dy}{dx}+2y=1$

$\Rightarrow y_1 (x)=Ae^{-2x}+\frac{1}{2}\mbox{ where }A\mbox{ is an arbitrary constant.}$

When $$x>3$$ we have,

$\frac{dy}{dx}+2y=0$

$\Rightarrow y_2 (x)=Be^{-2x}\mbox{ where }B\mbox{ is an arbitrary constant.}$

Since the solution should be a continuous function,

$y_1 (3)=\lim_{x\rightarrow 3^+}y_2 (x)$

$\Rightarrow y_1 (3)=y_2 (3)$

$\Rightarrow B=A+\frac{e^6}{2}$

Therefore the solution for the differential equation is,

$y(x)=\begin{cases}Ae^{-2x}+\frac{1}{2}&\mbox{ when } 0\leq x\leq 3\\Ae^{-2x}+\frac{e^{-2x+6}}{2} &\mbox{ when } x>3\end{cases}$

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