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

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Chris L T521

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Jan 26, 2012
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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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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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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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