The solution of the differential equation (dy)/(dx) - y + 3x _ e ( y + 3x ) + 3 = 0 is (where C is a constant of integration)

Correct answer: A. x - (1)/(2) ( _ e ( y + 3x )) ^ 2 = C (dy)/(dx) - (y + 3x)/( ( y + 3x )) + 3 = 0 (dy)/(dx) + 3 = (y + 3x)/( ( y + 3x )) (d)/(dx) ( y + 3x ) = (y + 3x)/( ( y + 3x )) ∫ ( ( y + 3x ))/(( y + 3x ))d ( y + 3x ) = ∫ dx Let ( y + 3x ) = t (1)/(( y + 3x ))d ( y + 3x ) = dt ∫ tdt = ∫ dx t ^ 2 2 = x + c ( ( y + 3x ) ) ^ 2 2 = x + c

JEE Main 20204 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 4 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2020 · 4 September · Evening Shift · Mathematics · Differential Equations

The solution of the differential equation (dy)/(dx) - y + 3x _ e (...

The solution of the differential equation

\dfrac{dy}{dx} - \dfrac{y + 3x}{\log_{e} ( y + 3x )} + 3 = 0

is
(where

C

is a constant of integration)

Options

A $x - \dfrac{1}{2}{\left(\log_{e} ( y + 3x )\right)}^{2} = C$
B $x - \log_{e} ( y + 3x ) = C$
C $y + 3x - \dfrac{1}{2}{\left(\log_{e}x\right)}^{2} = C$
D $x - 2\log_{e} ( y + 3x ) = C$

Correct answer

A. $x - \dfrac{1}{2}{\left(\log_{e} ( y + 3x )\right)}^{2} = C$

Step-by-step solution

\dfrac{dy}{dx} - \dfrac{y + 3x}{\ln ( y + 3x )} + 3 = 0

\dfrac{dy}{dx} + 3 = \dfrac{y + 3x}{\ln ( y + 3x )}

\dfrac{d}{dx} ( y + 3x ) = \dfrac{y + 3x}{\ln ( y + 3x )}

∫ \dfrac{\ln ( y + 3x )}{( y + 3x )}d ( y + 3x ) = ∫ dx

Let

\ln ( y + 3x ) = t

\dfrac{1}{( y + 3x )}d ( y + 3x ) = dt

∫ tdt = ∫ dx

\dfrac{{t}^{2}}{2} = x + c

\dfrac{( \ln ( y + 3x ) {)}^{2}}{2} = x + c

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