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The solution of $\frac{d y}{d x}+\frac{1}{3} y=1$ is
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1518 Upvotes
Verified Answer
The correct answer is:
$y=3+c e^{-x / 3}$
We have,
$$
\begin{array}{rlrl}
& \quad \frac{d y}{d x}+\frac{y}{3} & =1 \\
\Rightarrow & & d y & =-\frac{(y-3)}{3} d x \\
\Rightarrow & \frac{d y}{y-3} & =-\frac{1}{3} d x
\end{array}
$$
$\begin{aligned} & \Rightarrow \quad \log (y-3)=-\frac{1}{3} x+\log c \\ & \Rightarrow \quad y-3=c e^{-x / 3} \Rightarrow y=3+c e^{-x / 3}\end{aligned}$
$$
\begin{array}{rlrl}
& \quad \frac{d y}{d x}+\frac{y}{3} & =1 \\
\Rightarrow & & d y & =-\frac{(y-3)}{3} d x \\
\Rightarrow & \frac{d y}{y-3} & =-\frac{1}{3} d x
\end{array}
$$
$\begin{aligned} & \Rightarrow \quad \log (y-3)=-\frac{1}{3} x+\log c \\ & \Rightarrow \quad y-3=c e^{-x / 3} \Rightarrow y=3+c e^{-x / 3}\end{aligned}$
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