Question:
The solution of the differential equation
\(\frac{dy}{dx} - \frac{y+3x}{\log_e (y+3x)} + 3 = 0\)
is:
(where c is a constant of integration)
The solution of the differential equation
\(\frac{dy}{dx} - \frac{y+3x}{\log_e (y+3x)} + 3 = 0\)
is:
(where c is a constant of integration)
\(\frac{dy}{dx} - \frac{y+3x}{\log_e (y+3x)} + 3 = 0\)
is:
(where c is a constant of integration)
Show Hint
Substitute \(v = y + 3x\) so that \(\frac{dv}{dx} = \frac{dy}{dx} + 3\); the variables then separate.
Updated On: Nov 5, 2026
- \(x - \frac12\left(\log_e (y+3x)\right)^2 = C\)
- \(y + 3x - \frac12\left(\log_e x\right)^2 = C\)
- \(x - \log_e (y+3x) = C\)
- \(x - 2\log_e (y+3x) = C\)
Show Solution
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The Correct Option is A
Solution and Explanation
Let \(v = y + 3x\). Then \(\frac{dv}{dx} = \frac{dy}{dx} + 3\).
The equation becomes \(\frac{dv}{dx} = \frac{v}{\log_e v}\).
Separate: \(\frac{\log_e v}{v}\,dv = dx\).
Integrate: \(\frac12(\log_e v)^2 = x + c\).
So \(x - \frac12\left(\log_e(y+3x)\right)^2 = C\) with \(C = -c\).
Answer: (1) \(x - \frac12\left(\log_e (y+3x)\right)^2 = C\)
The equation becomes \(\frac{dv}{dx} = \frac{v}{\log_e v}\).
Separate: \(\frac{\log_e v}{v}\,dv = dx\).
Integrate: \(\frac12(\log_e v)^2 = x + c\).
So \(x - \frac12\left(\log_e(y+3x)\right)^2 = C\) with \(C = -c\).
Answer: (1) \(x - \frac12\left(\log_e (y+3x)\right)^2 = C\)
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