The particular solution of \(\rm \log \frac{d y}{d x}=3 x+4 y\), y(0) = 0 is

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  1. e3x + 3e−4y = 4
  2. 4e3x − 3−4y = 3
  3. 3e3x + 4e4y = 7
  4. 4e3x + 3e−4y = 7 

Answer (Detailed Solution Below)

Option 4 : 4e3x + 3e−4y = 7 
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Detailed Solution

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Calculation

log \(\frac{dy}{dx}\) = 3x + 4y

\(\frac{dy}{dx}\) = e3x+4y

dy ⋅ e-4y = e3x dx

\(\int\) e-4y dy = \(\int\) e3x dx

\(\frac{e^{-4y}}{-4}\) = \(\frac{e^{3x}}{3}\) + c

when x = 0, y = 0

\(\frac{1}{-4}\) = \(\frac{1}{3}\) + c

c = \(\frac{-7}{12}\)

\(\frac{e^{-4y}}{-4}\) = \(\frac{e^{3x}}{3}\) - \(\frac{7}{12}\)

-3e-4y = 4e3x - 7

4e3x + 3e-4y = 7

Hence option 4 is correct

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