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What is the solution of \(\frac{d y}{d x}+2 y=1\) satisfying \(y(0)=0\) ?
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Verified Answer
The correct answer is:
\(y=\frac{1-e^{-2 x}}{2}\)
\(\begin{aligned}
& \frac{d y}{d x}+2 y=1 \Rightarrow \frac{d y}{d x}=1-2 y \\
& \int \frac{d y}{1-2 y}=\int d x-\frac{1}{2} \log |1-2 y|=x+C \\
& \text { at } x=0, y=0 ;-\frac{1}{2} \log 1=0+C \Rightarrow C=0 \\
& 1-2 y=e^{-2 x} \Rightarrow y=\frac{1-e^{-2 x}}{2}
\end{aligned}\)
& \frac{d y}{d x}+2 y=1 \Rightarrow \frac{d y}{d x}=1-2 y \\
& \int \frac{d y}{1-2 y}=\int d x-\frac{1}{2} \log |1-2 y|=x+C \\
& \text { at } x=0, y=0 ;-\frac{1}{2} \log 1=0+C \Rightarrow C=0 \\
& 1-2 y=e^{-2 x} \Rightarrow y=\frac{1-e^{-2 x}}{2}
\end{aligned}\)
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