JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Differential Equations

The solution of differential equation

\frac{d y}{d x} - \frac{y + 3 x}{l o g _{e} ( y + 3 x )} + 3 = 0

is: (where C is a constant of integration.)

Answer: (B)

Solution

Writing the equation as

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

and putting

v = y + 3 x

, so that

\frac{d v}{d x} = \frac{d y}{d x} + 3

, turns it into

\frac{d v}{d x} = \frac{v}{ln v}

, that is

\frac{ln v}{v} d v = d x

. With

t = ln v

the left side integrates to

\frac{t ^{2}}{2}

, so

\frac{( ln ( y + 3 x ) ) ^{2}}{2} = x + C

, which rearranges to

x - \frac{1}{2} (ln (y + 3 x))^{2} = C

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